Trigonometric Functions: Class 11 Maths Chapter 3 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

100 min
1 simulation
59 practice questions
28 MCQs

Is chapter mein aap kya seekhenge

Angles ko rotation ki tarah degrees aur radians mein naapo, dono units ke beech convert karo, aur arcs ke liye l = rθ use karo.
Unit circle ki madad se kisi bhi real number ke six trigonometric functions define karo aur unhe quadrantal aur standard angles par evaluate karo.
Trigonometric functions ke signs, domains, ranges aur periods nikalo aur unke graphs sketch karo.
Ek value aur quadrant se saari six values nikalo, aur bade positive ya negative angles par functions evaluate karo.
Sum aur difference formulas, allied-angle results aur double- aur triple-angle formulas prove karo aur unhe apply karo.
Expressions simplify karne aur identities prove karne ke liye sum-to-product aur product-to-sum formulas use karo.

NCERT Mathematics Class 11, Chapter 3, pp. 43–75 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

3.1 Introduction

Dussehra ke mele mein giant wheel ki ek seat par nazar rakho. Wheel ek steady rate se ghoomta hai aur seat upar jaati hai, neeche aati hai aur phir upar jaati hai, aur har chakkar mein heights ka yahi pattern repeat hota hai. Ek right triangle ise describe nahi kar sakta. Seat jaise hi top se aage nikalti hai, uska turn is se zyada ho jaata hai: $90^{\circ},$ aur kuch chakkaron ke baad is se bhi zyada: $1000^{\circ} .$

5/5Ek full turn ek poori wave banata hai; agla turn use repeat karta hai.

Fig. 3.1Ghoomte wheel par ek seat ki height ek repeat hone wali wave banati hai.

Left: ek stand par giant wheel, jismein ek highlighted spoke par ek seat dot se marked hai. Right: axes jismein height upar ki taraf aur ghooma hua angle aage ki taraf hai. Step 1: seat hub ke level par hai aur graph ka point origin par hai. Step 2: quarter turn ke baad seat top par hai aur wave ka pehla quarter upar uthkar peak tak pahunchta hai. Step 3: half turn ke baad seat wapas hub ke level par hai aur wave wapas axis par aa jaati hai. Step 4: three quarters turn ke baad seat bottom par hai aur wave apne sabse neeche wale point tak pahunchti hai. Step 5: full turn ke baad seat wahin wapas aa jaati hai jahan se chali thi aur ek poori wave ban chuki hai.

Jo bhi cheez ghoomti hai, ya aage-peeche jhoolti hai, aisa hi pattern banati hai. Indian gharon ka 50-hertz alternating current, Mumbai mein tide ka chadhna aur utarna, vibrate karti sitar ki string aur Earth ke chakkar lagata satellite, sabko is chapter ke functions se describe kiya jaata hai. Inka parent subject, trigonometry (त्रिकोणमिति), apne naam mein abhi bhi apna purana kaam rakhta hai: iske peeche ke Greek words ka matlab hai triangle aur measure.

Tumhare Class 10 wale tools seat ko zyada der tak follow nahi kar sakte. Wahan $\sin θ$ ek right triangle mein opposite over hypotenuse tha, jo tabhi tak sense banata hai jab tak $θ$ acute hai. Jo kuch kaam karta tha, woh sab hum rakhenge, jaise $30^{\circ},$ $45^{\circ}$ aur $60^{\circ}$ par values aur identity $\sin^{2} θ + \cos^{2} θ = 1,$ aur angle par lagi limit hata denge. Plan yeh hai: angles ko turning ki amount ki tarah naapo (degrees mein aur ek naye unit, radian, mein), ek circle se har real number ke liye sine aur cosine define karo, unke graphs banao, aur do angles ke sum ka sine aur cosine nikalo.

Quick check

2 mein se 1

5 m lambi ek ladder deewar se tiki hai. Uska foot deewar se 3 m door hai, aur woh ground ke saath angle

θ

banati hai.

\cos θ

kya hai?

Answer dekho

Answer 0.6

\cos θ = \frac{adjacent}{hypotenuse} = \frac{3}{5} = 0.6.

Deewar ki 4 m height opposite side hai, jo iske bajaye

\sin θ = 0.8

deti hai.

3.2 Angles

Hinge par ghoomte ek darwaze ko imagine karo. Is chapter mein angle isi tarah turning ki ek amount hai: ek ray ek position se start hoti hai, jise uski initial side kehte hain, aur apne endpoint, yaani vertex, ke around tab tak ghoomti hai jab tak woh apni terminal side tak na pahunch jaaye. Ek sign direction record karta hai: anticlockwise turning ke liye positive aur clockwise ke liye negative. Ray ko ek se zyada baar ghoomne se koi nahi rokta, isliye ek angle ek full turn se bada bhi ho sakta hai (Fig. 3.2).

2/4 $60^{\circ}$ ka anticlockwise turn ek positive angle deta hai.

Fig. 3.2Angle turning ki ek amount hai, jiske saath direction ke liye ek sign hota hai.NCERT Fig. 3.1 · p. 43

Ek fixed horizontal ray jispar initial side likha hai, vertex O naam ke point se start hoti hai; ek doosri, coloured ray O ke around ghoomti hai. Step 1: ghoomne wali ray initial side par padi hai. Step 2: woh 60 degrees anticlockwise ghoom chuki hai, ek curved arrow par plus 60 degrees likha hai aur note hai anticlockwise turn: positive. Step 3: iske bajaye woh 45 degrees clockwise ghoomi hai, arrow par minus 45 degrees likha hai aur note hai clockwise turn: negative. Step 4: ek spiral arrow ek baar poora ghoomkar aage 60 degrees tak jaata hai, 420 degrees marked hai, aur note hai ek full turn, phir 60 degrees aur.

Turning ko poore turns, yaani revolutions, mein bhi gina ja sakta hai; motors aur fans isi tarah rate kiye jaate hain, revolutions per minute mein, aur full speed par ek ceiling fan kuch sau revolutions per minute karta hai. Zyadatar geometry ke liye poore turns bahut mote hain, isliye do baareek units use hote hain: degree aur radian.

3.2.1 Degree measure

Ek degree (अंश), likhne mein $1^{\circ},$ ek full turn ka $\frac{1}{360}$ hissa hai. Isse bhi baareek hain minute, yaani degree ka saathvaan hissa, aur second, yaani minute ka saathvaan hissa:

1^{\circ} = 60^{'}, 1^{'} = 60^{''}

Isliye full turn hai $360^{\circ},$ half turn $180^{\circ}$ aur quarter turn $90^{\circ} .$ $450^{\circ}$ ka angle ek full turn ke baad ek quarter turn hai, jabki $- 135^{\circ}$ clockwise ek turn ka teen-aathvaan hissa hai. Minutes aur seconds abhi bhi astronomy aur map-making mein aate hain; jaise, Earth ka axis apne orbit ke perpendicular se lagbhag $23^{\circ} 26^{'}$ jhuka hua hai.

3.2.2 Radian measure

Number 360 insaanon ki choice hai, circle ke liye iska koi khaas matlab nahi. Circle se hi liya gaya unit zyada useful hai. Radius jitni lambi ek string ko rim ke along bichhao (Fig. 3.3). Centre par yeh jo angle cover karti hai, woh har circle ke liye same hota hai, chahe circle bada ho ya chhota, aur use ek radian kehte hain.

Definition

Radian

Circle ke centre par woh angle jo radius jitni lambi arc banati hai. Unit circle (radius 1) par length 1 ki arc 1 radian mark karti hai.

ExampleRadius 5 cm wale circle par 5 cm lambi arc centre par theek 1 radian subtend karti hai.

2/4Rim ke along bichhane par yeh centre par 1 radian ka angle mark karti hai.

Fig. 3.3Ek radian woh angle hai jo radius jitni lambi arc banati hai.NCERT Fig. 3.4 · p. 45

Centre O wala ek circle aur ek radius OA jispar r likha hai. Step 1: same length r ki ek seedhi string A par khadi hai. Step 2: string ko A se rim ke along bichhaya gaya hai, uska end O se joda gaya hai; shaded sector ka angle 1 marked hai, notes ke saath arc equals r aur theta equals 1 radian. Step 3: do guna lambi string ek angle cover karti hai jo 2 marked hai, saath mein arc equals 2r aur theta equals 2 radians. Step 4: rim ke around 1, 2, 3, 4, 5 aur 6 radians par tick marks hain aur thoda gap bach jaata hai, aur note kehta hai ki rim ke around lagbhag 6.28 radii fit hoti hain, isliye ek full turn 2 pi radians hai.

Circumference hai $2 π r,$ isliye rim ke around theek $2 π$ radius-lengths fit hoti hain, aur ek complete turn $2 π$ radians ka hota hai. Centre par angle apni arc ke saath barabar ratio mein badhta hai, isliye length $l$ ki arc, jo $l / r$ radius-lengths lambi hai, $l / r$ radians subtend karti hai. Yahi arc-length formula hai:

Arc length

l = r θ

(3.1)

jahan

$l$: arc ki length
$r$: circle ka radius
$θ$: centre par angle, radians mein

3.2.3 Radian aur real numbers ka relation

Radians ki madad se hum angles ko simple real numbers ki tarah treat kar sakte hain. Centre $O$ ke saath unit circle banao aur yeh point mark karo: $A = (1, 0) .$ $A$ par ek number line aise khadi karo ki woh circle ko touch kare, $A$ par $0$ ho aur positive numbers upar ki taraf jaayein (Fig. 3.4). Ab positive half ko circle ke around anticlockwise lapeto aur negative half ko clockwise, jaise spool par dhaaga lapet-te hain.

4/4Number 2, $O$ par 2 radians ka angle mark karta hai.

Fig. 3.4Real line ko unit circle ke around lapetne se har real number radians mein ek angle ke saath jud jaata hai.NCERT Fig. 3.5 · p. 46

Centre O wala ek unit circle aur ek vertical number line jo use A par touch karti hai, jahan line par 0 likha hai; 1 aur 2 upar hain, minus 1 aur minus 2 neeche. Step 1: sirf seedhi number line dikhti hai. Step 2: upar wala half fade ho jaata hai aur uski copy circle ke around anticlockwise lipti hai, rim par points 1 aur 2 marked hain aur A ke saamne wale point par pi marked hai. Step 3: neeche wala half clockwise lipta hai, axis ke neeche minus 1 aur minus 2 marked hain. Step 4: radii O ko A se aur point 2 se jodti hain, unke beech ka angle marked hai, aur note kehta hai ki real number 2, 2 radians ka angle mark karta hai.

Har real number $t$ circle ke ek point par pahunchta hai, aur $A$ se us point tak ki arc ki length $∣ t ∣$ hai. Isliye $t$ , $t$ radians ka angle mark karta hai: positive numbers anticlockwise angles dete hain aur negative numbers clockwise. Alag numbers same point par pahunch sakte hain, jaise $t$ aur $t + 2 π,$ lekin har real number theek ek angle ka radian measure hai aur har angle ka ek radian measure hota hai. Isi wajah se hum sine aur cosine mein koi bhi real number daal sakte hain.

3.2.4 Degree aur radian ka relation

Ek full turn $360^{\circ}$ hai aur $2 π$ radians bhi. Dono ko aadha karne par woh ek fact milta hai jo har conversion ke liye chahiye:

Degrees aur radians

π rad = 180^{\circ}

(3.2)

Divide karne par, $1$ rad $= {(\frac{180}{π})}^{\circ} \approx {57.2958}^{\circ},$ yaani lagbhag $57^{\circ} 17^{'} 45^{''},$ aur $1^{\circ} = \frac{π}{180}$ rad, yaani lagbhag $0.01745$ rad. Rough value $π \approx \frac{22}{7}$ lene par pehla figure aata hai lagbhag $57^{\circ} 16^{'},$ jo zyadatar problems ke liye kaafi close hai. Convert karne ke liye sahi fraction se multiply karo:

Convert karna

\begin{aligned} y^{\circ} & = \frac{π y}{180} rad \\ x rad & = {(\frac{180 x}{π})}^{\circ} \end{aligned}

(3.3)

Table 3.1Common angles dono units mein

Degrees	Radians	Turn ka fraction
$30^{\circ}$	$\frac{π}{6}$	$\frac{1}{12}$
$45^{\circ}$	$\frac{π}{4}$	$\frac{1}{8}$
$60^{\circ}$	$\frac{π}{3}$	$\frac{1}{6}$
$90^{\circ}$	$\frac{π}{2}$	$\frac{1}{4}$
$120^{\circ}$	$\frac{2 π}{3}$	$\frac{1}{3}$
$135^{\circ}$	$\frac{3 π}{4}$	$\frac{3}{8}$
$150^{\circ}$	$\frac{5 π}{6}$	$\frac{5}{12}$
$180^{\circ}$	$π$	$\frac{1}{2}$
$270^{\circ}$	$\frac{3 π}{2}$	$\frac{3}{4}$
$360^{\circ}$	$2 π$	$1$

Example 1NCERT Example 1 · p. 47

Degrees aur minutes se radians

$47^{\circ} 15^{'}$ ko radians mein convert karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2NCERT Example 2 · p. 47

Radians se degrees, minutes aur seconds

5 radians ko degrees, minutes aur seconds mein convert karo, yeh lekar: $π = \frac{22}{7} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3NCERT Example 3 · p. 48

Arc se radius

Ek gol pizza ke slice ka tip angle $45^{\circ}$ hai aur uska crust 11 cm lamba hai. Pizza ka radius kya hai? Lo $π = \frac{22}{7} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4NCERT Example 4 · p. 48

Wiper blade ka tip

Ek windscreen wiper apne pivot se tip tak 42 cm lamba hai aur itna ghoomkar sweep karta hai: $120^{\circ} .$ Ek sweep mein tip kitni door chalta hai? Lo $π = \frac{22}{7} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5NCERT Example 5 · p. 48

Same arc, do circles

Ek park mein do gol flower beds ke kinaron par 3 m ki ek jaisi metal strips lagi hain. Ek strip apne bed ke centre par $40^{\circ}$ banati hai, doosri yeh: $72^{\circ} .$ Dono beds ke radii compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

150^{\circ}

ko radians mein

k π

ki tarah likho aur

k

type karo (3/4 jaisa fraction bhi chalega).

Answer dekho

Answer 0.8333 π

150 \times \frac{π}{180} = \frac{5 π}{6},

isliye

k = \frac{5}{6} .

Practice · 9 questions

Practice 3.1: answers ke saath

1
Short answerEasyNCERT Exercise 3.1 Q1
Radians mein convert karo.
1. $35^{\circ}$
  Solution dikhao
  Answer $\frac{7 π}{36}$
  $35 \times \frac{π}{180} = \frac{7 π}{36}$
2. $- 62^{\circ} 30^{'}$
  Solution dikhao
  Answer $- \frac{25 π}{72}$
  $- 62 \frac{1}{2} \times \frac{π}{180} = - \frac{125 π}{360}$
  $= - \frac{25 π}{72}$
3. $225^{\circ}$
  Solution dikhao
  Answer $\frac{5 π}{4}$
  $225 \times \frac{π}{180} = \frac{5 π}{4}$
4. $600^{\circ}$
  Solution dikhao
  Answer $\frac{10 π}{3}$
  $600 \times \frac{π}{180} = \frac{10 π}{3}$
2
Short answerMediumNCERT Exercise 3.1 Q2
Degrees mein convert karo (aur jahan zaroori ho wahan minutes aur seconds mein), yeh lekar: $π = \frac{22}{7} .$
1. $\frac{11}{24}$ rad
  Solution dikhao
  Answer $26^{\circ} 15^{'}$
  $\frac{11}{24} \times \frac{630}{11} = 26 {\frac{1}{4}}^{\circ}$
  Degree ka ek quarter $15^{'}$ hota hai.
2. $- 3$ rad
  Solution dikhao
  Answerlagbhag $- 171^{\circ} 49^{'} 5^{''}$
  $- 3 \times \frac{630}{11} = - 171 {\frac{9}{11}}^{\circ}$
  $\frac{9}{11} \times 60^{'} = 49 {\frac{1}{11}}^{'}$
  aur $\frac{1}{11} \times 60^{''} \approx 5^{''} .$
3. $\frac{4 π}{3}$
  Solution dikhao
  Answer $240^{\circ}$
  $\frac{4 π}{3} \times \frac{180}{π} = 240^{\circ}$
4. $\frac{11 π}{6}$
  Solution dikhao
  Answer $330^{\circ}$
  $\frac{11 π}{6} \times \frac{180}{π} = 330^{\circ}$
NumericalEasyNCERT Exercise 3.1 Q3
Ek grinder disc 450 revolutions per minute se ghoomti hai. Har second mein yeh kitna angle (radians mein) sweep karti hai, nikalo.
Solution dikhao
Answer $15 π$ rad, lagbhag $47.1$ rad
$\frac{450}{60} = 7.5 rev/s$
$7.5 \times 2 π = 15 π \approx 47.1 rad$
NumericalEasyNCERT Exercise 3.1 Q4
Ek railway track radius 1050 m ki circular arc ke along mudta hai. Is track ke 330 m par train ki direction kitne degrees ghoomti hai? Lo $π = \frac{22}{7} .$
°
Solution dikhao
Answer $18^{\circ}$
Travel ki direction (tangent) utna hi ghoomti hai jitna radius, yaani centre par bana angle.
$θ = \frac{330}{1050} = \frac{11}{35} rad$
$\frac{11}{35} \times \frac{630}{11} = 18^{\circ}$
NumericalMediumNCERT Exercise 3.1 Q5
Radius 9 cm wale circle mein ek chord $9 \sqrt{3}$ cm lambi hai. Chord se kati minor arc ki length nikalo.
Solution dikhao
Answer $6 π$ cm, lagbhag $18.8$ cm
Chord ka aadha $\frac{9 \sqrt{3}}{2}$ hai aur radius 9 hai, isliye central angle ke aadhe $α$ ke liye
$\sin α = \frac{9 \sqrt{3} / 2}{9} = \frac{\sqrt{3}}{2}$
Isliye $α = 60^{\circ}$ aur central angle hai $120^{\circ} = \frac{2 π}{3} .$
$l = 9 \times \frac{2 π}{3} = 6 π c m$
Short answerEasyNCERT Exercise 3.1 Q6
Ek jeweller gold wire ke do barabar tukdon ko alag-alag size ki do bangles ke along modta hai. Centre par naapne par ek tukda apni bangle ka $50^{\circ}$ cover karta hai, aur doosra yeh: $80^{\circ} .$ Bangles ke radii ka ratio nikalo.
Solution dikhao
Answer8 : 5
$\frac{r_{1}}{r_{2}} = \frac{θ_{2}}{θ_{1}} = \frac{80}{50} = \frac{8}{5}$
Radians wala conversion factor cancel ho jaata hai, isliye ratio mein seedhe degrees use kar sakte hain.
7
NumericalEasyNCERT Exercise 3.1 Q7
Playground ka ek jhoola 2.4 m lambi chains par latka hai. Jab seat di gayi length ki arc ke along chalti hai, tab chains kitne angle (radians mein) se ghoomti hain, nikalo.
1. 0.6 m
  rad
  Answer dekho
  Answer 0.25 rad
  $θ = \frac{0.6}{2.4} = \frac{1}{4}$ rad.
2. 1.2 m
  rad
  Answer dekho
  Answer 0.5 rad
  $θ = \frac{1.2}{2.4} = \frac{1}{2}$ rad.
3. 2 m
  rad
  Answer dekho
  Answer 0.8333 rad
  $θ = \frac{2}{2.4} = \frac{5}{6}$ rad.
Quick answerMedium
4:20 par clock ki suiyon ke beech kitna angle hai? Ise $k π$ radians ki tarah likho aur $k$ type karo.
π
Answer dekho
Answer 0.0556 π
Minute wali sui 12 se $120^{\circ}$ par hai. Hour wali sui yahan hai: $4 \times 30^{\circ} + 20 \times {0.5}^{\circ} = 130^{\circ} .$ Angle hai $10^{\circ} = \frac{π}{18},$ isliye $k = \frac{1}{18} .$
Quick answerMedium
Radius 35 cm ka ek wheel bina slip kiye road par 11 m roll karta hai. Yeh kitne radians ghoomta hai?
rad
Answer dekho
Answer 31.43 rad
$θ = \frac{l}{r} = \frac{1100}{35} = \frac{220}{7} \approx 31.4$ rad, jo 5 full turns hai jab $π$ ki value yeh lete hain: $\frac{22}{7} .$

3.3 Trigonometric Functions

Ek right triangle ko aise rakho ki uska acute angle $x$ radius 1 wale circle ke centre par ho aur uska hypotenuse ek radius ho. Hypotenuse 1 hai, isliye $\cos x$ (adjacent over hypotenuse) circle par wale corner ke horizontal coordinate ke barabar hai, aur $\sin x$ uske vertical coordinate ke barabar. Isi observation se pata chalta hai ki jab koi triangle hi na ho tab sine aur cosine kaise define karein: unhe circle se padh lo.

Maan lo unit circle ka centre $O$ origin par hai, aur $A = (1, 0) .$ Kisi bhi real number $x$ ke liye, $A$ se circle ke around length $x$ ki arc chalo, anticlockwise agar $x > 0$ aur clockwise agar $x < 0,$ aur jis point par pahuncho uska naam rakho: $P (a, b) .$ Tab angle $A O P$ , $x$ radians hota hai (Fig. 3.5).

Definition

Real number ka sine aur cosine

Agar unit circle par $A (1, 0)$ se arc length $x$ wala point yeh hai: $P (a, b),$ to $\cos x = a$ aur $\sin x = b .$

ExampleArc length $\frac{π}{2}$ par point yeh hai: $(0, 1),$ isliye $\cos \frac{π}{2} = 0$ aur $\sin \frac{π}{2} = 1.$

3/4 $P$ ke coordinates ke naam hain $\cos x$ aur $\sin x .$

Fig. 3.5Sine aur cosine unit circle par ek point ke coordinates hain.NCERT Fig. 3.6 · p. 49

Origin O par centred radius 1 ke circle ke saath axes, aur point A (1, 0) par. Step 1: length x ki ek coloured arc A se first quadrant ke ek point P tak anticlockwise jaati hai. Step 2: P se ek perpendicular x-axis ko M par milta hai, aur right triangle OMP shaded hai jiski sides hain a axis ke along, b upar ki taraf aur hypotenuse 1; panel par likha hai a squared plus b squared equals 1. Step 3: panel coordinates ka naam badalta hai: cos x equals a aur sin x equals b, isliye cos squared x plus sin squared x equals 1. Step 4: triangle fade ho jaata hai aur chaar dots A, B, C aur D mark karte hain jahan circle axes se milta hai, angles 0, pi by 2, pi aur 3 pi by 2 par.

Triangle $O M P$ mein $M$ par right angle hai, saath mein $O M = ∣ a ∣,$ $M P = ∣ b ∣$ aur $O P = 1.$ Pythagoras se circle ke har point ke liye, har quadrant mein, $a^{2} + b^{2} = 1$ milta hai, kyunki square karne se signs hat jaate hain. Naye notation mein:

Pythagorean identity

\cos^{2} x + \sin^{2} x = 1

(3.4)

Yahan $\cos^{2} x$ ka matlab hai $(\cos x)^{2} .$ Yeh identity har real $x$ ke liye sahi hai, sirf acute angles ke liye nahi.

Fig. 3.5 ka last step woh chaar points mark karta hai jahan circle axes se milta hai. Yahan is ke whole-number multiples se pahunchte hain: $\frac{π}{2},$ isliye unhe quadrantal angles kehte hain, aur unke sines aur cosines seedhe padhe ja sakte hain. Pair $(\cos x, \sin x)$ ki values yeh hain: $(1, 0)$ jab $x = 0,$ $(0, 1)$ jab $x = \frac{π}{2},$ $(- 1, 0)$ jab $x = π$ aur $(0, - 1)$ jab $x = \frac{3 π}{2}$ ; $x = 2 π$ par hum wapas yahan hote hain: $(1, 0) .$

Length $x + 2 π$ ki arc, length $x$ ki arc ke baad ek extra lap hai, aur lap wahin khatam hota hai jahan se shuru hua tha. Yahi baat kitne bhi whole laps ke liye sahi hai, aage ya peeche, isliye

Poore turns se kuch nahi badalta

\begin{array}{c} \sin (x + 2 n π) = \sin x \\ \cos (x + 2 n π) = \cos x \end{array}

(3.5)

jahan

$n$: koi bhi integer

Sine $P$ ki height hai, isliye yeh theek tab zero hota hai jab $P$ , $A$ ya $C$ par ho, yaani jab $x$ , $π$ ka multiple ho. Cosine theek tab zero hota hai jab $P$ , $B$ ya $D$ par ho, yaani jab $x$ , $\frac{π}{2}$ guna koi odd integer ho:

Sine aur cosine ke zeros

\begin{array}{c} \sin x = 0 \Leftrightarrow x = n π \\ \cos x = 0 \Leftrightarrow \\ x = (2 n + 1) \frac{π}{2} \end{array}

(3.6)

jahan

$n$: koi bhi integer

Baaki chaar functions sine aur cosine se bante hain. Har ek wahan defined hai jahan uska denominator zero nahi hai:

Baaki chaar functions

\begin{aligned} \tan x & = \sin x / \cos x \\ \cot x & = \cos x / \sin x \\ \sec x & = 1 / \cos x \\ cosec x & = 1 / \sin x \end{aligned}

(3.7)

jahan

$\tan x, \sec x$: defined jab $x \neq (2 n + 1) \frac{π}{2}$
$\cot x, cosec x$: defined jab $x \neq n π$

Do aur Pythagorean identities turant nikal aati hain. (3.4) ko is se divide karo: $\cos^{2} x,$ jo tab allowed hai jab $\cos x \neq 0$ ; phir iske bajaye ise is se divide karo: $\sin^{2} x,$ jo tab allowed hai jab $\sin x \neq 0$ :

Aur Pythagorean identities

\begin{array}{c} 1 + \tan^{2} x = \sec^{2} x \\ 1 + \cot^{2} x = {cosec}^{2} x \end{array}

(3.8)

$0 < x < \frac{π}{2}$ ke liye triangle $O M P$ bilkul Class 10 wala right triangle hai, isliye nayi definitions purane ratios se match karti hain aur jaani-pehchaani values waise hi rehti hain (Table 3.2). Sine column yaad rakhne ka ek aasaan tareeka: yeh padhta hai $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2},$ aur cosine column yahi list ulte order mein hai.

Table 3.2First quadrant ke standard angles par values

$x$	$\sin x$	$\cos x$	$\tan x$
$0$	$0$	$1$	$0$
$\frac{π}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
$\frac{π}{4}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	$1$
$\frac{π}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$\frac{π}{2}$	$1$	$0$	defined nahi

Reciprocal functions ke liye entries ulat do: jaise $\sec \frac{π}{3} = 2$ aur $\cot \frac{π}{6} = \sqrt{3},$ jabki $cosec 0$ exist nahi karta.

Quick check

3 mein se 1

\sin \frac{3 π}{2}

kya hai?

1

0

defined nahi

- 1

Answer dekho

Answer (D) $- 1$

Three quarters turn yahan khatam hota hai:

D (0, - 1),

aur sine doosra coordinate hai, yaani

- 1

. Answer

0

yeh hai:

\cos \frac{3 π}{2},

yaani pehla coordinate.

3.3.1 Trigonometric functions ke signs

Arc length $x$ wale point $P (a, b)$ ko arc length $- x$ wale point $Q$ se compare karo. Dono arcs barabar lambi hain lekin $A$ se ulti directions mein chalti hain, isliye $Q$ , $x$ -axis mein $P$ ki mirror image hai: $Q = (a, - b)$ (Fig. 3.6). Yeh har quadrant mein kaam karta hai, isliye saare real $x$ ke liye:

Negative angles

\begin{array}{c} \cos (- x) = \cos x \\ \sin (- x) = - \sin x \end{array}

(3.9)

3/3Pehla coordinate same, doosra ulta: cosine even hai, sine odd hai.

Fig. 3.6Arc ki direction ulat dene se point

x

-axis mein reflect ho jaata hai.NCERT Fig. 3.7 · p. 51

Axes ke saath ek unit circle. Step 1: x label wali ek anticlockwise arc A se second quadrant ke point P(a, b) tak jaati hai. Step 2: minus x label wali, utni hi lambi ek clockwise arc A se third quadrant ke point Q(a, minus b) tak jaati hai, aur ek dashed vertical line x-axis ke aar-paar P aur Q ko jodti hai. Step 3: panel par likha hai pehla coordinate same, doosra ulta, uske baad cos of minus x equals cos x aur sin of minus x equals minus sin x.

Jis function ke liye har $x$ par $f (- x) = f (x)$ ho, woh even function hai; jiske liye $f (- x) = - f (x)$ ho, woh odd function hai. Isliye cosine even hai aur sine odd. Dono rules ko divide karne par milta hai $\tan (- x) = - \tan x,$ isliye tangent bhi odd hai.

Kyunki $P$ kabhi unit circle se bahar nahi jaata, isliye koi bhi coordinate size mein 1 se zyada nahi ho sakta: har $x$ ke liye $∣ \sin x ∣ \leq 1$ aur $∣ \cos x ∣ \leq 1$ . Dono coordinates ke signs sirf is par depend karte hain ki $P$ kis quadrant mein hai, aur baaki chaar functions sine aur cosine se bane hain, isliye unke signs bhi wahin se aate hain. Fig. 3.7 chaaron quadrants ka chakkar lagata hai.

4/4IV mein $P = (+, -)$ : sirf cosine aur secant positive hain.

Fig. 3.7Quadrants I se IV tak positive functions A, S, T, C banate hain (yaad rakho Add Sugar To Coffee).NCERT p. 52

Axes plane ko chaar tiles mein baantte hain, aur origin se ek ghoomne wala arm current quadrant ki taraf point karta hai. Step 1: quadrant I highlighted hai aur usmein letter A, words saare 6 aur sign pair plus, plus dikhte hain. Step 2: arm quadrant II mein ghoomta hai, jismein S, sin aur cosec, aur minus, plus dikhte hain. Step 3: quadrant III mein T, tan aur cot, aur minus, minus dikhte hain. Step 4: quadrant IV mein C, cos aur sec, aur plus, minus dikhte hain; ab chaaron tiles labelled hain.

Table 3.3Quadrant ke hisaab se six functions ke signs

Quadrant	Positive	Negative
I	saare six	koi nahi
II	sin, cosec	cos, sec, tan, cot
III	tan, cot	sin, cosec, cos, sec
IV	cos, sec	sin, cosec, tan, cot

Yahi reasoning kisi bhi radius $r$ wale circle par kaam karti hai. Agar $P (a, b)$ us circle par hai aur $O P$ positive $x$ -axis ke saath angle $x$ banata hai, to $\cos x = \frac{a}{r}$ aur $\sin x = \frac{b}{r},$ kyunki triangle $O M P$ unit-circle wala triangle hi hai, bas $r$ guna bada. Simulation mein liya gaya hai $r = 5,$ jahan $(3, 4)$ aur $(- 4, 3)$ jaise grid points exact fractions dete hain.

SimulationRadius 5 wale circle par points

Dono axes par minus 6 se 6 tak ka coordinate grid, chaaron quadrants labelled. Point O origin par fixed hai aur point P (3, 4) se start hota hai; distance OP draw karke naapi gayi hai, aur x-axis mein P ki mirror image dikhayi gayi hai. P ko drag karne par woh grid points par snap hota hai aur uske coordinates aur distance update hote hain.

Fig. 3.8

P

ko aise grid points par drag karo jahan

O P = 5

ho, aur use

x

-axis mein uski mirror image se compare karo.

Try karo

$P$ ko yahan le jao: $(- 4, 3) .$ Check karo ki $O P = 5,$ phir nikalo ki $\cos x = - \frac{4}{5}$ aur $\sin x = \frac{3}{5} .$ Yeh kaun sa quadrant hai?
$P$ ki mirror image padho. $\cos x$ aur $\sin x$ mein se kiska sign badalta hai, aur kaun same rehta hai?
$P$ ko pehle yahan rakho: $(0, - 5),$ phir yahan: $(- 5, 0) .$ In quadrantal angles par $\sin x$ aur $\cos x$ kya hain?

Quick check

3 mein se 1

Agar

\sin x < 0

aur

\tan x > 0,

x

kis quadrant mein khatam hota hai?

IIIIIIIV

Answer dekho

Answer (C) III

Sine III aur IV mein negative hai; tangent I aur III mein positive hai. Sirf III fit hota hai. Quadrant IV chunne ka mann karta hai, lekin wahan tangent negative hai.

3.3.2 Trigonometric functions ka domain aur range

Sine aur cosine har real number ke liye defined hain, kyunki har $x$ circle par ek point deta hai, aur unki values us point ke coordinates hain, isliye dono ki range hai $[- 1, 1] .$ Baaki chaar functions woh points kho dete hain jahan unke denominators zero ho jaate hain. Kyunki $[- 1, 1]$ ke kisi number ka reciprocal is interval ke bahar hota hai: $(- 1, 1),$ isliye secant aur cosecant kabhi strictly $- 1$ aur $1$ ke beech ki values nahi lete. Table 3.4 mein saare results hain; $n$ koi bhi integer hai.

Table 3.4Domains aur ranges

Function	Domain	Range
$\sin x$	$ℝ$	$[- 1, 1]$
$\cos x$	$ℝ$	$[- 1, 1]$
$\tan x$	$ℝ - {(2 n + 1) \frac{π}{2}}$	$ℝ$
$\cot x$	$ℝ - {n π}$	$ℝ$
$\sec x$	$ℝ - {(2 n + 1) \frac{π}{2}}$	$(- \infty, - 1] \cup [1, \infty)$
$cosec x$	$ℝ - {n π}$	$(- \infty, - 1] \cup [1, \infty)$

Jaise $x$ , $0$ se $2 π$ tak chalta hai, point $P$ circle ka ek chakkar lagata hai, aur tum padh sakte ho ki har function kaise badalta hai. Height $\sin x$ quadrant I mein $0$ se $1$ tak chadhti hai, II mein wapas $0$ par girti hai, III mein $- 1$ tak utarti hai aur IV mein wapas $0$ par aati hai. Cosine $1$ se start hota hai, $x = π$ par lagataar girkar $- 1$ tak pahunchta hai aur phir wapas $1$ tak chadhta hai. Tangent har quadrant mein badhta hai: $0$ se shuru hokar jaise $x$ , $\frac{π}{2}$ ke paas aata hai, yeh bina kisi limit ke badhta jaata hai, $\tan x \to \infty$ likhna isi bina limit wale badhne ka short form hai. $\infty$ koi number nahi hai, aur $\tan x$ kabhi iske barabar nahi hota. phir $\frac{π}{2}$ ke theek baad bahut bade negative values se wapas aata hai aur phir se badhta hai.

Ek function $f$ periodic hai agar koi number $p > 0$ aisa ho ki uske domain ke har $x$ ke liye $f (x + p) = f (x)$ ; aisa sabse chhota $p$ uska period hai. Sine aur cosine ka period $2 π$ hai, aur unke reciprocals ka bhi. Tangent aur cotangent jaldi repeat hote hain, $π$ ke baad: half turn $P (a, b)$ ko yahan bhejta hai: $(- a, - b),$ jisse ratio $\frac{b}{a}$ nahi badalta. Hum $\tan (π + x) = \tan x$ ko Section 3.4 mein prove karenge.

Values plot karne par Fig. 3.9, Fig. 3.10 aur Fig. 3.11 wale graphs milte hain. Jahan function undefined hota hai, wahan uska graph ek vertical line ke saath-saath chalta hai, uske paas aur paas aata hai lekin kabhi use touch nahi karta. Aisi line ko asymptote kehte hain.

2/4 $y = \cos x$ (dashed) ka shape same hai, aur peak yahan hai: $x = 0.$

Fig. 3.9Sine aur cosine waves: same shape, bas quarter turn ka shift.NCERT Figs 3.8, 3.9 · p. 54

Minus pi se 3 pi tak axes, heights 1 aur minus 1 par grid lines ke saath. Step 1: sine curve, origin se guzarti ek smooth wave jo pi by 2 par peak karti hai aur 3 pi by 2 par minus 1 tak neeche jaati hai. Step 2: cosine curve, dashed, jodi jaati hai; iska shape same hai lekin peak 0 aur 2 pi par hai. Step 3: 0 par cosine peak se pi by 2 par sine peak tak ek arrow hai jispar pi by 2 likha hai. Step 4: axis ke neeche 0 se 2 pi tak ek double arrow hai jispar likha hai ek period equals 2 pi.

2/3 $y = \cot x$ (dashed) $π$ ke multiples par bane asymptotes ke beech girta hai.

Fig. 3.10Tangent aur cotangent har

π

par repeat hote hain aur unke vertical asymptotes hote hain.NCERT Figs 3.10, 3.11 · p. 54

Minus pi se 2 pi tak axes, heights lagbhag 3 tak. Step 1: tangent curve teen branches mein, har branch minus pi by 2, pi by 2 aur 3 pi by 2 par bane dashed vertical asymptotes ke beech bahut neeche se bahut upar tak chadhti hai, aur 0 aur pi par axis cross karti hai. Step 2: tangent fade hota hai aur cotangent dikhta hai, dashed, jo minus pi, 0, pi aur 2 pi par bane asymptotes ke beech har branch mein girta hai. Step 3: sirf tangent, 0 se pi tak ek double arrow ke saath jispar period pi likha hai.

3/3Koi bhi function strictly $- 1$ aur $1$ ke beech ki values nahi leta.

Fig. 3.11Secant aur cosecant kabhi strictly

- 1

aur

1

ke beech ki values nahi lete.NCERT Figs 3.12, 3.13 · p. 55

Minus pi se 2 pi tak axes. Step 1: cosine curve dotted hai, aur secant curve U-shaped branches mein bana hai jo cosine curve ko uske peaks aur troughs par touch karti hain aur pi by 2 ke odd multiples par bane dashed asymptotes ke saath upar ya neeche jaati hain. Step 2: secant ki jagah dotted sine curve aur cosecant branches aati hain, jo sine curve ko uske peaks aur troughs par touch karti hain, aur asymptotes pi ke multiples par hain. Step 3: heights minus 1 aur 1 ke beech ka ek band shaded hai jispar likha hai koi value nahi.

InteractiveUnit circle lab

Ek slider minus 720 se 720 degrees tak angle set karta hai, saath mein buttons hain 15 degrees ke steps ke liye, ek full turn sweep karne ke liye, aur angle ko radians mein dikhane ke liye. Ek unit circle point P dikhata hai, A se P tak ki arc (extra full turns ke liye ek dashed ring aur note ke saath), x-axis ke along ek solid blue segment ki tarah cosine aur upar ki taraf ek dashed red segment ki tarah sine; tangent mode mein A par tangent line par ek green segment tan x dikhata hai. Iske bagal mein minus 4 pi se 4 pi tak sine, cosine ya tangent ka graph current angle mark karta hai. Ek panel quadrant ka naam aur sin, cos aur tan ke signs batata hai, 30 aur 45 degrees ke multiples par exact values deta hai, aur batata hai ki 0 aur 360 degrees ke beech kaun sa angle same point par pahunchta hai.

Fig. 3.12Angle ghumao aur dekho ki

P

, uske coordinates aur graph saath-saath kaise chalte hain.

Upar ki descriptions ko test karne ke liye lab use karo. Ek full turn sweep karne ke liye Play dabao aur dekho ki dot sine curve kaise trace karta hai. Phir graph ko tangent par switch karo aur tangent line par green segment ko dekho: angle jaise jaise is ke paas aata hai, woh tezi se lamba hota jaata hai: $90^{\circ} .$

Signs aur teen Pythagorean identities ke saath, kisi trigonometric function ki ek value aur quadrant se baaki saari values nikal sakte hain.

Example 6NCERT Example 6 · p. 55

Ek sine se paanch values

Diya hai $\sin x = - \frac{8}{17}$ aur $x$ quadrant III mein hai; baaki paanch trigonometric functions nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7NCERT Example 7 · p. 56

Ek tangent se paanch values

Diya hai $\tan x = - \frac{20}{21}$ aur $x$ quadrant II mein hai; baaki paanch trigonometric functions nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8NCERT Example 8 · p. 56

Radians mein ek bada angle

Yeh nikalo: $\cos \frac{37 π}{6} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 57

Degrees mein ek bada negative angle

Yeh nikalo: $\sin (- 1830^{\circ}) .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Inmein se kaun si

\sec x

ki possible value hai?

0

\frac{1}{2}

- \frac{2}{3}

- \frac{3}{2}

Answer dekho

Answer (D) $- \frac{3}{2}$

∣ \sec x ∣ = \frac{1}{∣ \cos x ∣} \geq 1,

isliye sirf

- \frac{3}{2}

ho sakta hai. Values

\frac{1}{2}

aur

- \frac{2}{3}

is ke liye possible hain:

\cos x,

lekin uske reciprocal ke liye nahi.

Practice · 11 questions

Practice 3.2: answers ke saath

Short answerEasyNCERT Exercise 3.2 Q1
Diya hai $\cos x = - \frac{35}{37}$ aur $x$ quadrant III mein hai; baaki paanch functions nikalo.
Solution dikhao
Answer $\sin x = - \frac{12}{37},$ $\tan x = \frac{12}{35},$ $\cot x = \frac{35}{12},$ $\sec x = - \frac{37}{35},$ $cosec x = - \frac{37}{12}$
$\sin^{2} x = 1 - \frac{1225}{1369} = \frac{144}{1369}$
Quadrant III mein sine negative hota hai, isliye $\sin x = - \frac{12}{37} .$
$\tan x = \frac{- 12}{- 35} = \frac{12}{35}$
Baaki teen reciprocals hain.
Short answerEasyNCERT Exercise 3.2 Q2
Diya hai $\sin x = \frac{7}{25}$ aur $x$ quadrant II mein hai; baaki paanch functions nikalo.
Solution dikhao
Answer $\cos x = - \frac{24}{25},$ $\tan x = - \frac{7}{24},$ $\cot x = - \frac{24}{7},$ $\sec x = - \frac{25}{24},$ $cosec x = \frac{25}{7}$
$\cos^{2} x = 1 - \frac{49}{625} = \frac{576}{625}$
Quadrant II mein cosine negative hota hai, isliye $\cos x = - \frac{24}{25} .$
$\tan x = \frac{7}{- 24} = - \frac{7}{24}$
Short answerMediumNCERT Exercise 3.2 Q3
Diya hai $\cot x = \frac{40}{9}$ aur $x$ quadrant III mein hai; baaki paanch functions nikalo.
Solution dikhao
Answer $\tan x = \frac{9}{40},$ $\sin x = - \frac{9}{41},$ $\cos x = - \frac{40}{41},$ $cosec x = - \frac{41}{9},$ $\sec x = - \frac{41}{40}$
${cosec}^{2} x = 1 + \frac{1600}{81} = \frac{1681}{81}$
Quadrant III mein sine negative hota hai, isliye $cosec x = - \frac{41}{9}$ aur $\sin x = - \frac{9}{41} .$
$\begin{aligned} \cos x & = \cot x \sin x \\ = \frac{40}{9} (- \frac{9}{41}) = - \frac{40}{41} \end{aligned}$
Short answerEasyNCERT Exercise 3.2 Q4
Diya hai $\sec x = \frac{61}{11}$ aur $x$ quadrant IV mein hai; baaki paanch functions nikalo.
Solution dikhao
Answer $\cos x = \frac{11}{61},$ $\sin x = - \frac{60}{61},$ $\tan x = - \frac{60}{11},$ $\cot x = - \frac{11}{60},$ $cosec x = - \frac{61}{60}$
$\cos x = \frac{11}{61}$
$\sin^{2} x = 1 - \frac{121}{3721} = \frac{3600}{3721}$
Quadrant IV mein sine negative hota hai, isliye $\sin x = - \frac{60}{61}$ aur $\tan x = - \frac{60}{11} .$
Short answerMediumNCERT Exercise 3.2 Q5
Diya hai $\tan x = - \frac{45}{28}$ aur $x$ quadrant II mein hai; baaki paanch functions nikalo.
Solution dikhao
Answer $\sec x = - \frac{53}{28},$ $\cos x = - \frac{28}{53},$ $\sin x = \frac{45}{53},$ $cosec x = \frac{53}{45},$ $\cot x = - \frac{28}{45}$
$\sec^{2} x = 1 + \frac{2025}{784} = \frac{2809}{784}$
Quadrant II mein secant negative hota hai: $\sec x = - \frac{53}{28},$ isliye $\cos x = - \frac{28}{53} .$
$\sin x = \tan x \cos x = \frac{45}{53}$
Quick answerEasyNCERT Exercise 3.2 Q6
Yeh nikalo: $\cos 1125^{\circ} .$
Answer dekho
Answer 0.7071
$1125^{\circ} - 3 \times 360^{\circ} = 45^{\circ},$ isliye value hai $\cos 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.7071.$
Quick answerEasyNCERT Exercise 3.2 Q7
Yeh nikalo: $\sec (- 1380^{\circ}) .$
Answer dekho
Answer 2
$- 1380^{\circ} + 4 \times 360^{\circ} = 60^{\circ},$ isliye value hai $\sec 60^{\circ} = 2.$
Quick answerEasyNCERT Exercise 3.2 Q8
Yeh nikalo: $\tan \frac{25 π}{4} .$
Answer dekho
Answer 1
$\frac{25 π}{4} = 6 π + \frac{π}{4},$ isliye value hai $\tan \frac{π}{4} = 1.$
Quick answerEasyNCERT Exercise 3.2 Q9
Yeh nikalo: $\cos (- \frac{17 π}{3}) .$
Answer dekho
Answer 0.5
$- \frac{17 π}{3} + 6 π = \frac{π}{3},$ isliye value hai $\cos \frac{π}{3} = \frac{1}{2} .$
Quick answerEasyNCERT Exercise 3.2 Q10
Yeh nikalo: $cosec (- \frac{23 π}{6}) .$
Answer dekho
Answer 2
$- \frac{23 π}{6} + 4 π = \frac{π}{6},$ isliye value hai $cosec \frac{π}{6} = 2.$
Quick answerMedium
Assertion (A): Kisi bhi real number $x$ ke liye yeh nahi hota: $\sec x = \frac{1}{2} .$ Reason (R): Har real $x$ ke liye $∣ \cos x ∣ \leq 1$ .
A aur R dono true hain, aur R, A ko explain karta hai.A aur R dono true hain, lekin R, A ko explain nahi karta.A true hai lekin R false hai.A false hai lekin R true hai.
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ko explain karta hai.
$\sec x = \frac{1}{2}$ ke liye chahiye hoga $\cos x = 2,$ jise R rule out karta hai. Isliye A true hai, R true hai, aur R hi exact reason hai.

3.4 Do Angles ke Sum aur Difference ke Trigonometric Functions

$\sin x$ aur $\sin y$ jaan lene se addition karke $\sin (x + y)$ nahi milta: $\sin 30^{\circ} + \sin 60^{\circ} \approx 1.37,$ jabki $\sin 90^{\circ} = 1.$ Sahi rules, yaani sum aur difference formulas, poore course ki sabse useful trigonometric identities hain. Do identities Section 3.3.1 se pehle hi hamare paas hain: $\sin (- x) = - \sin x$ aur $\cos (- x) = \cos x .$ Is section mein baaki sab kuch ek aur identity se nikalta hai, yaani is formula se: $\cos (x - y) .$

Difference aur sum ka cosine

2/3 $y$ jitna clockwise ghumane par $Q$ , $A$ par aa jaata hai aur $P$ , $R$ par, jiski arc length hai $x - y .$

Fig. 3.13Chord ko centre ke around ghumane se uski length nahi badalti.NCERT Fig. 3.14 · p. 58

Centre O wala ek unit circle. Step 1: circle par arc lengths x aur y par points P aur Q, jinke labels P(cos x, sin x) aur Q(cos y, sin y) hain, O tak radii, chord PQ moti line se bani hai, aur radii ke beech angle x minus y marked hai. Step 2: triangle OPQ angle y jitna clockwise ghoomta hai; Q, A(1, 0) par aa jaata hai aur P, R par, jo arc length x minus y wala point hai, aur purani chord ki ek dotted copy peeche reh jaati hai. Step 3: note par likha hai barabar chords: O par same angle, aur PQ equals RA.

Unit circle par lo $P (\cos x, \sin x)$ aur $Q (\cos y, \sin y),$ taaki radii $O P$ aur $O Q$ , $O$ par angle $x - y$ banayein (Fig. 3.13). Distance formula se, $\cos^{2} + \sin^{2} = 1$ ko do baar use karke, milta hai

\begin{aligned} P Q^{2} \\ = (\cos x - \cos y)^{2} \\ + (\sin x - \sin y)^{2} \\ = 2 - 2 \cos x \cos y \\ - 2 \sin x \sin y \end{aligned}

Ab triangle $O P Q$ ko $O$ ke around angle $y$ jitna clockwise ghumao. $Q$ yahan pahunchta hai: $A (1, 0),$ aur $P$ , $R$ par pahunchta hai, jiski arc length hai $x - y,$ isliye $R = (\cos (x - y), \sin (x - y)) .$ Ghumane se lengths nahi badalti, isliye $R A = P Q,$ aur

\begin{aligned} R A^{2} \\ = (\cos (x - y) - 1)^{2} \\ + \sin^{2} (x - y) \\ = 2 - 2 \cos (x - y) \end{aligned}

Dono expressions ko barabar rakhkar aur cancel karke result milta hai, jo saare real $x$ aur $y$ ke liye valid hai:

Difference ka cosine

\begin{aligned} \cos (x - y) \\ = \cos x \cos y \\ + \sin x \sin y \end{aligned}

(3.10)

$y$ ki jagah $- y$ rakhne aur yeh use karne par: $\cos (- y) = \cos y$ aur $\sin (- y) = - \sin y,$ sum formula milta hai:

Sum ka cosine

\begin{aligned} \cos (x + y) \\ = \cos x \cos y \\ - \sin x \sin y \end{aligned}

(3.11)

Sum ka sine, aur allied angles

(3.10) mein $x = \frac{π}{2}$ rakho. Kyunki $\cos \frac{π}{2} = 0$ aur $\sin \frac{π}{2} = 1,$ isliye milta hai $\cos (\frac{π}{2} - y) = \sin y .$ Is result mein $y$ ki jagah $\frac{π}{2} - y$ rakhne par milta hai $\sin (\frac{π}{2} - y) = \cos y .$ Yeh co-function rules hain: kisi angle ka sine uske complement ka cosine hota hai.

Co-function rules

\begin{array}{c} \cos (\frac{π}{2} - x) = \sin x \\ \sin (\frac{π}{2} - x) = \cos x \end{array}

(3.12)

Co-function rules ek sine ko cosine mein badal dete hain, isliye sum ka sine (3.10) se nikal aata hai. Kaise, yeh dekhne ke liye steps kholo.

Sum ya difference ka sine

\begin{aligned} \sin (x \pm y) \\ = \sin x \cos y \\ \pm \cos x \sin y \end{aligned}

(3.13)

In formulas mein $x$ ya $y$ ko quadrantal angle chunne par $\frac{π}{2} + x,$ $π \pm x$ aur $2 π - x$ par values, $x$ par values ke terms mein mil jaati hain. Aise angles ko allied angles kehte hain. Jaise, $\sin (π - x) = \sin π \cos x - \cos π \sin x = \sin x .$ Fig. 3.14 ki pictures har case ko yaad rakhna aasaan bana deti hain, kyunki har allied point is point ka reflection ya turn hai: $P (a, b) .$

5/5Quarter turn se $\frac{π}{2} + x$ wala point milta hai: $(- b, a) .$

Fig. 3.14Allied angles, yaani is point ke reflections aur turns:

P (a, b) .

NCERT p. 59

First quadrant mein ek chhote angle x par point P(a, b) wala ek unit circle. Step 1: sirf P, saath mein sin x equals b aur cos x equals a. Step 2: y-axis ek dashed mirror ki tarah bana hai aur P reflect hokar pi minus x par (minus a, b) ban jaata hai, saath mein sin of pi minus x equals sin x aur cos of pi minus x equals minus cos x. Step 3: O se guzarti ek dashed line P ko pi plus x par (minus a, minus b) se jodti hai, yaani half turn. Step 4: x-axis mirror hai aur image 2 pi minus x par (a, minus b) hai. Step 5: ek curved arrow pi by 2 plus x par (minus b, a) tak quarter turn dikhata hai, saath mein sin of pi by 2 plus x equals cos x aur cos of pi by 2 plus x equals minus sin x.

Table 3.5x ke terms mein allied angles

Angle	sin	cos	tan
$\frac{π}{2} - x$	$\cos x$	$\sin x$	$\cot x$
$\frac{π}{2} + x$	$\cos x$	$- \sin x$	$- \cot x$
$π - x$	$\sin x$	$- \cos x$	$- \tan x$
$π + x$	$- \sin x$	$- \cos x$	$\tan x$
$2 π - x$	$- \sin x$	$\cos x$	$- \tan x$

Signs ka ek quick test: maan lo $x$ ek chhota acute angle hai, dekho ki allied angle kis quadrant mein hai, aur answer ko wahi sign do jo us quadrant mein function ka hota hai.

Sum ka tangent aur cotangent

Maan lo $x$ , $y$ aur $x + y,$ teenon ke cosines mein se koi bhi zero nahi hai. $\tan (x + y)$ ko sine aur cosine ke quotient ki tarah likho, dono ko expand karo, aur upar aur neeche dono ko is se divide karo: $\cos x \cos y .$ Details ke liye steps kholo.

Sum ya difference ka tangent

\begin{aligned} \tan (x \pm y) \\ = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y} \end{aligned}

(3.14)

Cotangent bhi isi tarah kaam karta hai. Agar $x$ , $y$ aur $x + y$ ke sines sab non-zero hon, to iske bajaye $\sin x \sin y$ se divide karo aur milega

Sum ya difference ka cotangent

\begin{aligned} \cot (x \pm y) \\ = \frac{\cot x \cot y \mp 1}{\cot y \pm \cot x} \end{aligned}

(3.15)

Example 10NCERT Example 10 · p. 64

Standard angles par evaluate karna

Dikhao ki $3 \tan \frac{π}{4} cosec \frac{π}{6} - 4 \cos \frac{2 π}{3} \sec \frac{π}{3} = 10.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 11NCERT Example 11 · p. 64

Sum se ek exact value

Iski exact value nikalo: $\cos 105^{\circ} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12NCERT Example 12 · p. 64

π se aage ka tangent

Yeh nikalo: $\tan \frac{19 π}{12} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13NCERT Example 13 · p. 65

Tangents mein ek quotient

Prove karo ki un saare $x$ aur $y$ ke liye jahan dono sides defined hain, $\frac{\cos (x + y)}{\cos (x - y)} = \frac{1 - \tan x \tan y}{1 + \tan x \tan y}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14NCERT Example 14 · p. 65

Teen tangents ka product

Dikhao ki jahan bhi teenon tangents defined hain, wahan $\tan 5 x \tan 3 x \tan 2 x = \tan 5 x - \tan 3 x - \tan 2 x$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

\sin 75^{\circ}

barabar hai

\frac{\sqrt{6} + \sqrt{2}}{4}

\frac{\sqrt{6} - \sqrt{2}}{4}

\frac{\sqrt{3} + 1}{2}

\sin 45^{\circ} + \sin 30^{\circ}

Answer dekho

Answer (A) $\frac{\sqrt{6} + \sqrt{2}}{4}$

\sin (45^{\circ} + 30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.966.

Sines ko seedha jodne par lagbhag

1.21

milta hai, jo 1 se zyada hai, aur koi bhi sine itna nahi ho sakta.

Double aur triple angles

Sum formulas mein $y = x$ rakhne par $2 x$ par values milti hain. Cosine ke liye, (3.4) ki madad se hum $\sin^{2} x$ aur $\cos^{2} x$ ko ek doosre se badal sakte hain, isliye kai barabar useful forms hain.

Double angle ka cosine

\begin{aligned} \cos 2 x \\ = \cos^{2} x - \sin^{2} x \\ = 2 \cos^{2} x - 1 \\ = 1 - 2 \sin^{2} x \\ = \frac{1 - \tan^{2} x}{1 + \tan^{2} x} \end{aligned}

(3.16)

Last form ke liye yeh likho: $\cos 2 x = \frac{\cos^{2} x - \sin^{2} x}{\cos^{2} x + \sin^{2} x},$ kyunki denominator 1 hai, aur upar aur neeche ko $\cos^{2} x$ se divide karo; iske liye chahiye $\cos x \neq 0.$ Yahi trick $\sin 2 x = 2 \sin x \cos x$ ko is ke formula mein badal deti hai: $\tan x .$

Double angle ka sine

\begin{aligned} \sin 2 x & = 2 \sin x \cos x \\ = \frac{2 \tan x}{1 + \tan^{2} x} \end{aligned}

(3.17)

Double angle ka tangent

\tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x}

(3.18)

Tangent formula (3.14) hi hai, bas yeh lekar: $y = x .$ Yeh tab sahi hai jab $\tan x$ defined ho aur $2 x$ is ka odd multiple na ho: $\frac{π}{2},$ taaki $\tan 2 x$ bhi exist kare.

(3.16) ke do rearrangements baar baar use hote hain, aksar $x$ ki jagah $\frac{x}{2}$ ke saath: $\cos^{2} x = \frac{1 + \cos 2 x}{2}$ aur $\sin^{2} x = \frac{1 - \cos 2 x}{2} .$ Yeh squares ko first powers mein badal dete hain, aur half-angle values dete hain (Example 21).

Sum formula ko ek baar aur lagao, is baar is par: $2 x + x,$ to triple-angle formulas milte hain.

\begin{aligned} \sin 3 x = \sin (2 x + x) \\ = \sin 2 x \cos x \\ + \cos 2 x \sin x \\ = 2 \sin x (1 - \sin^{2} x) \\ + \sin x - 2 \sin^{3} x \\ = 3 \sin x - 4 \sin^{3} x \end{aligned}

Teesri line mein use hua hai $\sin 2 x = 2 \sin x \cos x,$ $\cos^{2} x = 1 - \sin^{2} x$ aur $\cos 2 x = 1 - 2 \sin^{2} x .$ $\cos (2 x + x)$ aur $\tan (2 x + x)$ ko isi tarah expand karne par baaki do formulas milte hain; tangent wale ke liye zaroori hai ki $3 x$ , is ka odd multiple na ho: $\frac{π}{2} .$

Triple angles

\begin{aligned} \sin 3 x \\ = 3 \sin x - 4 \sin^{3} x \\ \cos 3 x \\ = 4 \cos^{3} x - 3 \cos x \\ \tan 3 x \\ = \frac{3 \tan x - \tan^{3} x}{1 - 3 \tan^{2} x} \end{aligned}

(3.19)

Quick check

3 mein se 1

Agar

\cos x = \frac{1}{3},

to yeh nikalo:

\cos 2 x .

Answer dekho

Answer -0.7778

\cos 2 x = 2 \cos^{2} x - 1 = \frac{2}{9} - 1 = - \frac{7}{9} \approx - 0.778.

Cosine ko double karna, yaani

\frac{2}{3},

woh galat answer hai jo sahi lagta hai.

Sums se products, products se sums

$\cos (x + y)$ aur $\cos (x - y)$ ke expansions jodo to sine terms cancel ho jaate hain; ghatao to cosine terms cancel ho jaate hain. Yahi in dono ke saath bhi hota hai: $\sin (x + y)$ aur $\sin (x - y) .$ Chaaron results product-to-sum formulas hain:

Product se sum

\begin{aligned} 2 \cos x \cos y \\ = \cos (x + y) \\ + \cos (x - y) \\ - 2 \sin x \sin y \\ = \cos (x + y) \\ - \cos (x - y) \\ 2 \sin x \cos y \\ = \sin (x + y) \\ + \sin (x - y) \\ 2 \cos x \sin y \\ = \sin (x + y) \\ - \sin (x - y) \end{aligned}

(3.20)

Ab inhe ulta padho. Rakho $x + y = C$ aur $x - y = D,$ taaki $x = \frac{C + D}{2}$ aur $y = \frac{C - D}{2} .$ Tab har formula ek sum ya difference ko product mein badal deta hai. $C$ aur $D$ ko phir se $x$ aur $y$ naam dene par sum-to-product formulas milte hain:

Sum se product

\begin{aligned} \cos x + \cos y \\ = 2 \cos \frac{x + y}{2} \cos \frac{x - y}{2} \\ \cos x - \cos y \\ = - 2 \sin \frac{x + y}{2} \sin \frac{x - y}{2} \\ \sin x + \sin y \\ = 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2} \\ \sin x - \sin y \\ = 2 \cos \frac{x + y}{2} \sin \frac{x - y}{2} \end{aligned}

(3.21)

Words mein: do cosines ka sum, half-sum ke cosine aur half-difference ke cosine ke product ka do guna hota hai, aur isi tarah baaki. Inka main use proofs mein hota hai, jahan bhi kisi sum ko factorise karna ho.

Yeh ek aisi sound ko bhi explain karte hain jo shayad tumne suni ho. Thodi alag frequencies wale do tuning forks ko ek saath bajao, to ek aisa note milta hai jo regular tarah se tez aur dheema hota hai; in swells ko beats kehte hain. Lagbhag barabar frequency ki do cosine waves jodne par ek fast wave milti hai jiska size ek slow cosine factor control karta hai (Fig. 3.15).

3/3Sum hai $2 \cos 6 x \cos x$ : slow factor $2 \cos x$ (dashed) envelope hai.

Fig. 3.15Do paas-paas ki frequencies jodne se beats bante hain.

Upar ka panel: do patli waves, cos 7x solid aur cos 5x dashed, 0 se 2 pi ke ek interval par; woh saath start hoti hain, step se bahar jaati hain aur wapas aa jaati hain. Step 2: neeche ka ek panel unka sum jodta hai, ek fast wave jiski height wahan 2 tak badhti hai jahan upar ki waves match karti hain aur wahan lagbhag 0 tak ghat jaati hai jahan match nahi karti. Step 3: do dashed curves, plus aur minus 2 cos x, sum ko upar aur neeche se gherti hain, label hai 2 cos 6x cos x.

Example 15NCERT Example 15 · p. 65

Do sines ka sum

Prove karo ki $\sin (\frac{π}{3} + x) + \sin (\frac{π}{3} - x) = \sqrt{3} \cos x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 16NCERT Example 16 · p. 66

Sums ka ek quotient

Prove karo ki $\frac{\sin 9 x + \sin 3 x}{\cos 9 x - \cos 3 x} = - \cot 3 x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 17NCERT Example 17 · p. 66

Factorise karne se pehle grouping

Prove karo ki $\frac{\sin 3 x + \sin x - 2 \sin 2 x}{\cos 3 x - \cos x} = \tan \frac{x}{2} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

\cos 75^{\circ} + \cos 15^{\circ}

barabar hai

\frac{\sqrt{3}}{2}

\sqrt{\frac{3}{2}}

\frac{1}{\sqrt{2}}

1

Answer dekho

Answer (B) $\sqrt{\frac{3}{2}}$

2 \cos 45^{\circ} \cos 30^{\circ} = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \sqrt{\frac{3}{2}} \approx 1.22.

Pehle angles jodna, yaani

\cos 90^{\circ} = 0,

allowed nahi hai.

Practice · 25 questions

Practice 3.3: answers ke saath

ProofEasyNCERT Exercise 3.3 Q1
Prove karo ki $\cos^{2} \frac{π}{6} + \sin^{2} \frac{π}{4} - \cot^{2} \frac{π}{4} = \frac{1}{4} .$
Solution dikhao
Yahan $\cos^{2} \frac{π}{6} = \frac{3}{4},$ $\sin^{2} \frac{π}{4} = \frac{1}{2}$ aur $\cot^{2} \frac{π}{4} = 1.$
$\frac{3}{4} + \frac{1}{2} - 1 = \frac{1}{4}$
ProofEasyNCERT Exercise 3.3 Q2
Prove karo ki $2 \cos^{2} \frac{π}{4} + \sec^{2} \frac{5 π}{6} \sin^{2} \frac{π}{6} = \frac{4}{3} .$
Solution dikhao
$\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2},$ isliye $\sec^{2} \frac{5 π}{6} = \frac{4}{3} .$
$2 \cdot \frac{1}{2} + \frac{4}{3} \cdot \frac{1}{4}$
$= 1 + \frac{1}{3} = \frac{4}{3}$
ProofEasyNCERT Exercise 3.3 Q3
Prove karo ki $\tan^{2} \frac{π}{3} + cosec \frac{7 π}{6} + 3 \cot^{2} \frac{π}{3} = 2.$
Solution dikhao
$\sin \frac{7 π}{6} = - \sin \frac{π}{6} = - \frac{1}{2},$ isliye $cosec \frac{7 π}{6} = - 2.$
$3 - 2 + 3 \cdot \frac{1}{3} = 2$
ProofEasyNCERT Exercise 3.3 Q4
Prove karo ki $2 \cos^{2} \frac{3 π}{4} + \tan^{2} \frac{2 π}{3} + 2 {cosec}^{2} \frac{π}{4} = 8.$
Solution dikhao
$\cos \frac{3 π}{4} = - \frac{1}{\sqrt{2}},$ $\tan \frac{2 π}{3} = - \sqrt{3},$ $cosec \frac{π}{4} = \sqrt{2} .$
$2 \cdot \frac{1}{2} + 3 + 2 \cdot 2 = 8$
5
Short answerMediumNCERT Exercise 3.3 Q5
Exact value nikalo
1. $\sin 105^{\circ}$
  Solution dikhao
  Answer $\frac{\sqrt{6} + \sqrt{2}}{4}$
  $\sin (60^{\circ} + 45^{\circ})$
  $= \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}}$
  $= \frac{\sqrt{6} + \sqrt{2}}{4}$
2. $\tan 165^{\circ}$
  Solution dikhao
  Answer $\sqrt{3} - 2$
  $\tan 165^{\circ} = \tan (180^{\circ} - 15^{\circ}) = - \tan 15^{\circ},$ aur
  $\tan 15^{\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$
  $= \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$
  isliye $\tan 165^{\circ} = \sqrt{3} - 2.$
ProofMediumNCERT Exercise 3.3 Q6
Prove karo ki $\cos (\frac{π}{3} + x) \cos (\frac{π}{6} + y) - \sin (\frac{π}{3} + x) \sin (\frac{π}{6} + y) = - \sin (x + y) .$
Solution dikhao
(3.11) se left side dono angles ke sum ka cosine hai:
$\cos (\frac{π}{2} + x + y)$
aur $\cos (\frac{π}{2} + θ) = - \sin θ,$ jisse milta hai $- \sin (x + y) .$
ProofMediumNCERT Exercise 3.3 Q7
Prove karo ki $\tan (\frac{π}{4} + x) - \tan (\frac{π}{4} - x) = 2 \tan 2 x .$
Solution dikhao
Maan lo $t = \tan x,$ to (3.14) se milta hai
$\frac{1 + t}{1 - t} - \frac{1 - t}{1 + t}$
$= \frac{(1 + t)^{2} - (1 - t)^{2}}{1 - t^{2}}$
$= \frac{4 t}{1 - t^{2}} = 2 \tan 2 x$
ProofMediumNCERT Exercise 3.3 Q8
Prove karo ki $\frac{\cos (\frac{π}{2} + x) \sec (- x)}{\sin (π - x) \cot (\frac{π}{2} - x)} = - cosec x .$
Solution dikhao
Upar hai $(- \sin x) \sec x = - \tan x,$ aur neeche hai $\sin x \tan x,$ isliye
$\frac{- \tan x}{\sin x \tan x} = - \frac{1}{\sin x}$
jo hai $- cosec x .$
ProofHardNCERT Exercise 3.3 Q9
Prove karo ki $\sin (\frac{3 π}{2} - x) \sin (2 π - x) [\tan (\frac{3 π}{2} + x) - \tan (π + x)] = - 1.$
Solution dikhao
$\sin (\frac{3 π}{2} - x) = - \cos x$ aur $\sin (2 π - x) = - \sin x$ ; inka product hai $\sin x \cos x .$
$\tan (\frac{3 π}{2} + x) = \frac{- \cos x}{\sin x} = - \cot x$ aur $\tan (π + x) = \tan x,$ isliye bracket hai
$- \frac{\cos^{2} x + \sin^{2} x}{\sin x \cos x}$
$= - \frac{1}{\sin x \cos x}$
$\sin x \cos x$ se multiply karne par $- 1$ milta hai.
ProofEasyNCERT Exercise 3.3 Q10
Prove karo ki $\sin (n + 3) x \cos (n + 1) x - \cos (n + 3) x \sin (n + 1) x = \sin 2 x .$
Solution dikhao
Yeh $\sin (A - B)$ hai jahan $A = (n + 3) x$ aur $B = (n + 1) x,$ aur $A - B = 2 x .$
ProofMediumNCERT Exercise 3.3 Q11
Prove karo ki $\cos (\frac{2 π}{3} + x) - \cos (\frac{2 π}{3} - x) = - \sqrt{3} \sin x .$
Solution dikhao
Half-sum $\frac{2 π}{3}$ hai aur half-difference $x$ hai:
$- 2 \sin \frac{2 π}{3} \sin x$
$= - 2 \cdot \frac{\sqrt{3}}{2} \sin x = - \sqrt{3} \sin x$
ProofMediumNCERT Exercise 3.3 Q12
Prove karo ki $\sin^{2} 5 x - \sin^{2} 2 x = \sin 7 x \sin 3 x .$
Solution dikhao
Yeh use karo: $\sin^{2} θ = \frac{1 - \cos 2 θ}{2}$ :
$\frac{1 - \cos 10 x}{2} - \frac{1 - \cos 4 x}{2}$
$= \frac{1}{2} (\cos 4 x - \cos 10 x)$
(3.21) se, $\cos 4 x - \cos 10 x = 2 \sin 7 x \sin 3 x,$ isliye left side hai $\sin 7 x \sin 3 x .$
ProofMediumNCERT Exercise 3.3 Q13
Prove karo ki $\cos^{2} x - \sin^{2} 3 x = \cos 4 x \cos 2 x .$
Solution dikhao
$\cos^{2} x = \frac{1 + \cos 2 x}{2}$ aur $\sin^{2} 3 x = \frac{1 - \cos 6 x}{2}$ use karo:
$LHS = \frac{1}{2} (\cos 6 x + \cos 2 x)$
$= \frac{1}{2} \cdot 2 \cos 4 x \cos 2 x$
$= \cos 4 x \cos 2 x$
ProofMediumNCERT Exercise 3.3 Q14
Prove karo ki $\sin x + 2 \sin 3 x + \sin 5 x = 4 \sin 3 x \cos^{2} x .$
Solution dikhao
$\sin 5 x + \sin x = 2 \sin 3 x \cos 2 x$
$LHS = 2 \sin 3 x (\cos 2 x + 1)$
aur $1 + \cos 2 x = 2 \cos^{2} x,$ isliye left side hai $4 \sin 3 x \cos^{2} x .$
ProofMediumNCERT Exercise 3.3 Q15
Prove karo ki $\tan 3 x (\cos 5 x + \cos x) = \sin 5 x + \sin x .$
Solution dikhao
$\begin{aligned} \cos 5 x + \cos x \\ = 2 \cos 3 x \cos 2 x \end{aligned}$
$\sin 5 x + \sin x = 2 \sin 3 x \cos 2 x$
Isliye $\tan 3 x \cdot 2 \cos 3 x \cos 2 x = 2 \sin 3 x \cos 2 x,$ jo right side hai.
ProofMediumNCERT Exercise 3.3 Q16
Prove karo ki $\frac{\cos 11 x - \cos 3 x}{\sin 15 x - \sin x} = - \frac{\sin 4 x}{\cos 8 x} .$
Solution dikhao
$\begin{aligned} \cos 11 x - \cos 3 x \\ = - 2 \sin 7 x \sin 4 x \end{aligned}$
$\begin{aligned} \sin 15 x - \sin x \\ = 2 \cos 8 x \sin 7 x \end{aligned}$
$2 \sin 7 x$ cancel karne par bachta hai $- \frac{\sin 4 x}{\cos 8 x} .$
ProofEasyNCERT Exercise 3.3 Q17
Prove karo ki $\frac{\sin 7 x - \sin x}{\cos 7 x + \cos x} = \tan 3 x .$
Solution dikhao
$\frac{2 \cos 4 x \sin 3 x}{2 \cos 4 x \cos 3 x} = \tan 3 x$
ProofMediumNCERT Exercise 3.3 Q18
Prove karo ki $\frac{\cos x - \cos y}{\sin x + \sin y} = \tan \frac{y - x}{2} .$
Solution dikhao
$\frac{- 2 \sin \frac{x + y}{2} \sin \frac{x - y}{2}}{2 \sin \frac{x + y}{2} \cos \frac{x - y}{2}}$
$= - \tan \frac{x - y}{2}$
aur $- \tan \frac{x - y}{2} = \tan \frac{y - x}{2}$ kyunki tangent odd hai.
ProofEasyNCERT Exercise 3.3 Q19
Prove karo ki $\frac{\sin 2 x + \sin 4 x}{\cos 2 x - \cos 4 x} = \cot x .$
Solution dikhao
$\frac{2 \sin 3 x \cos x}{- 2 \sin 3 x \sin (- x)}$
Kyunki $\sin (- x) = - \sin x,$ isliye yeh hai $\frac{\cos x}{\sin x} = \cot x .$
ProofEasyNCERT Exercise 3.3 Q20
Prove karo ki $\frac{\cos 3 x + \cos x}{\cos^{2} x - \sin^{2} x} = 2 \cos x .$
Solution dikhao
$\frac{2 \cos 2 x \cos x}{\cos 2 x} = 2 \cos x$
upar (3.21) aur neeche (3.16) use karke.
ProofMediumNCERT Exercise 3.3 Q21
Prove karo ki $\frac{\sin x + \sin 3 x + \sin 5 x}{\cos x + \cos 3 x + \cos 5 x} = \tan 3 x .$
Solution dikhao
$\sin 5 x + \sin x = 2 \sin 3 x \cos 2 x$
$\begin{aligned} \cos 5 x + \cos x \\ = 2 \cos 3 x \cos 2 x \end{aligned}$
$\frac{\sin 3 x (2 \cos 2 x + 1)}{\cos 3 x (2 \cos 2 x + 1)}$
jo hai $\tan 3 x .$
ProofHardNCERT Exercise 3.3 Q22
Prove karo ki $\cot x \cot 3 x - \cot 3 x \cot 4 x - \cot 4 x \cot x = 1.$
Solution dikhao
$\cot 4 x = \frac{\cot 3 x \cot x - 1}{\cot x + \cot 3 x}$
Cross-multiply karne par:
$\begin{aligned} \cot 4 x \cot x + \cot 4 x \cot 3 x \\ = \cot 3 x \cot x - 1 \end{aligned}$
Rearrange karne par result milta hai.
ProofHardNCERT Exercise 3.3 Q23
Prove karo ki $\cos 4 x = \frac{1 - 6 \tan^{2} x + \tan^{4} x}{(1 + \tan^{2} x)^{2}} .$
Solution dikhao
$t = \tan x$ aur $\tan 2 x = \frac{2 t}{1 - t^{2}}$ ke saath:
$\cos 4 x = \frac{1 - \tan^{2} 2 x}{1 + \tan^{2} 2 x}$
$= \frac{(1 - t^{2})^{2} - 4 t^{2}}{(1 - t^{2})^{2} + 4 t^{2}}$
Upar hai $1 - 6 t^{2} + t^{4}$ aur neeche hai $1 + 2 t^{2} + t^{4} = (1 + t^{2})^{2} .$
ProofMediumNCERT Exercise 3.3 Q24
Prove karo ki $\sin^{4} x + \cos^{4} x = \frac{3 + \cos 4 x}{4} .$
Solution dikhao
$\begin{aligned} (\sin^{2} x + \cos^{2} x)^{2} \\ - 2 \sin^{2} x \cos^{2} x \end{aligned}$
$= 1 - \frac{1}{2} \sin^{2} 2 x$
$= 1 - \frac{1}{4} (1 - \cos 4 x)$
$= \frac{3 + \cos 4 x}{4}$
ProofHardNCERT Exercise 3.3 Q25
Prove karo ki $\cos 5 x = 16 \cos^{5} x - 20 \cos^{3} x + 5 \cos x .$
Solution dikhao
Maan lo $c = \cos x .$ Tab $\cos 3 x = 4 c^{3} - 3 c$ aur $\cos 2 x = 2 c^{2} - 1,$ isliye
$\cos 5 x = 2 \cos 3 x \cos 2 x - c$
$= 2 (4 c^{3} - 3 c) (2 c^{2} - 1) - c$
$= 16 c^{5} - 20 c^{3} + 5 c$

Miscellaneous examples

Yeh examples chapter ke kai ideas ko milate hain: quadrant signs, sum formulas aur multiple-angle formulas. Har ek mein square root lene se pehle quadrant tay karo.

Example 18NCERT Example 18 · p. 68

Do angles ko milana

Agar $\cos x = - \frac{15}{17}$ aur $x$ second quadrant mein hai, aur $\sin y = - \frac{3}{5}$ aur $y$ fourth quadrant mein hai, to yeh nikalo: $\cos (x + y) .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 19NCERT Example 19 · p. 68

Products se sums aur wapas

Prove karo ki $\sin \frac{x}{2} \sin \frac{7 x}{2} + \sin \frac{3 x}{2} \sin \frac{11 x}{2} = \sin 2 x \sin 5 x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 20NCERT Example 20 · p. 69

Quadratic se ek exact tangent

Iski exact value nikalo: $\tan \frac{3 π}{8} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Half-angle questions mein pehle $\frac{x}{2}$ ki jagah pata karo: jo inequality $x$ ko uske quadrant mein rakhti hai, use aadha karne se pata chalta hai ki $\frac{x}{2}$ kahan hai, aur isliye kaun se signs chunne hain (Fig. 3.16).

3/3Isliye $\sin \frac{x}{2} > 0$ aur $\cos \frac{x}{2} < 0.$

Fig. 3.16Angle ko aadha karne se woh quadrant III se quadrant II mein aa jaata hai.NCERT p. 70

Axes aur quadrant labels ke saath ek unit circle. Step 1: quadrant III shaded hai aur A se ek arc anticlockwise pi aur 3 pi by 2 ke beech angle x wale point tak jaati hai. Step 2: woh arc fade hoti hai; pi by 2 se 3 pi by 4 tak ka sector shaded hai aur ek chhoti arc quadrant II mein angle x by 2 wale point tak jaati hai. Step 3: panel par likha hai x by 2, II mein hai, isliye x by 2 ka sine positive hai aur cosine negative.

Example 21NCERT Example 21 · p. 70

Half-angle values

Diya hai $\sin x = - \frac{24}{25}$ aur $π < x < \frac{3 π}{2},$ to yeh nikalo: $\cos \frac{x}{2},$ $\sin \frac{x}{2}$ aur $\tan \frac{x}{2} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 22NCERT Example 22 · p. 71

Teen sines ke squares

Prove karo ki $\sin^{2} x + \sin^{2} (x + \frac{2 π}{3}) + \sin^{2} (x - \frac{2 π}{3}) = \frac{3}{2} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Agar

\sin A = \frac{8}{17}

aur

\sin B = \frac{3}{5}

hain, aur

A

aur

B

acute hain, to

\sin (A + B)

ko decimal mein nikalo.

Answer dekho

Answer 0.9059

\cos A = \frac{15}{17}

aur

\cos B = \frac{4}{5},

isliye

\sin (A + B) = \frac{8}{17} \cdot \frac{4}{5} + \frac{15}{17} \cdot \frac{3}{5} = \frac{77}{85} \approx 0.906.

History se

Sine word kahan se aaya

Second century BCE ke Greek astronomer Hipparchus ne circles ki chords ki tables banayi. Indian astronomers ne iske bajaye half-chord ke saath kaam kiya, jo bilkul hamara sine hai, aur use ardha-jyā ya sirf jyā kaha, jiska matlab hai dhanush ki dori. Arabic scholars ne is word ko jība likha; baad mein ise jaib (ek fold ya pocket) padha gaya, aur Latin mein iska translation sinus hua, jo aage chalkar sine ban gaya.

Bhaskara I ne sine ko approximate karne ke liye ek kamaal ka accurate fraction diya, aur Bhaskara II ne sum aur difference ke sine ke rules bataye aur aise angles ke exact sines nikale, jaise $18^{\circ}$ aur $36^{\circ} .$ Kerala school ke Malayalam text Yuktibhāṣā ne aise results ke proofs diye, saath mein sine ki infinite series bhi. Bahut pehle, kaha jaata hai ki Miletus ke Thales ne ek pyramid ki height uski shadow ko ek stick ki shadow se compare karke nikali thi, yahi idea tumne Class 10 mein heights and distances ke liye use kiya tha.

Timeline

Milestones

c. 140 BCE
Hipparchus
Astronomy ke liye chords ki ek table.
499 CE
Aryabhata
Aryabhatiya mein half-chord (sine) ki ek table hai.
c. 629
Bhaskara I
Ek simple fraction jo sine ko kaafi achhe se approximate karta hai.
1150
Bhaskara II
Siddhānta Śiromaṇi, jismein sums ke sines aur exact values hain.
c. 1530
Yuktibhāṣā
Kerala school se proofs aur sine series, Malayalam mein.
1813
John Herschel
Inverse sine ke liye notation $\sin^{- 1} x$ introduce karte hain.

Recap

Summary aur key points

Angle turning ki ek amount hai: anticlockwise positive hai, clockwise negative (Section 3.2).
Ek degree ek turn ka $\frac{1}{360}$ hissa hai, saath mein $1^{\circ} = 60^{'}$ aur $1^{'} = 60^{''}$ (Section 3.2.1).
Radius ke barabar arc 1 radian subtend karti hai, aur $θ$ radians mein ho to $l = r θ$ (Section 3.2.2).
$π$ rad $= 180^{\circ}$ : convert karne ke liye $\frac{π}{180}$ se ya $\frac{180}{π}$ se multiply karo (Section 3.2.4).
Agar unit circle par arc length $x$ wala point yeh hai: $(a, b),$ to $\cos x = a,$ $\sin x = b$ aur $\cos^{2} x + \sin^{2} x = 1$ (Section 3.3).
$\cos^{2} x$ ya $\sin^{2} x$ se divide karne par baaki do Pythagorean identities milti hain ((3.8)).
Sine sirf $π$ ke multiples par zero hota hai, cosine sirf $\frac{π}{2}$ ke odd multiples par ((3.6)).
Cosine even hai; sine aur tangent odd hain. Quadrant ke hisaab se positive functions A, S, T, C banate hain (Section 3.3.1).
Sine, cosine, secant aur cosecant ka period $2 π$ hai; tangent aur cotangent ka period $π$ hai (Section 3.3.2).
Difference ka cosine unit circle ki ek chord se aata hai ((3.10)); sums aur differences ke cosine aur sine usi se nikalte hain ((3.11), (3.13)).
Allied angles $\frac{π}{2} \pm x,$ $π \pm x,$ $2 π - x$ sign ke fark ke saath $x$ wali values dete hain, aur $\frac{π}{2} \pm x$ par sine aur cosine aapas mein badal jaate hain (Sum ka sine, aur allied angles).
Sine formula ko cosine formula se divide karne par tangent rule milta hai ((3.14)); cotangent rule uski mirror hai ((3.15)).
$y = x$ rakhne par double angles milte hain ((3.16), (3.17), (3.18)); ek aur step se triple angles milte hain ((3.19)).
Sum formulas ko jodne aur ghatane se products sums mein badal jaate hain ((3.20)); unhe ulta padhne se sums factorise ho jaate hain ((3.21)).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 3.1 Q1	Practice 3.1 · Q1
NCERT Exercise 3.1 Q2	Practice 3.1 · Q2
NCERT Exercise 3.1 Q3	Practice 3.1 · Q3
NCERT Exercise 3.1 Q4	Practice 3.1 · Q4
NCERT Exercise 3.1 Q5	Practice 3.1 · Q5
NCERT Exercise 3.1 Q6	Practice 3.1 · Q6
NCERT Exercise 3.1 Q7	Practice 3.1 · Q7
NCERT Exercise 3.2 Q1	Practice 3.2 · Q1
NCERT Exercise 3.2 Q2	Practice 3.2 · Q2
NCERT Exercise 3.2 Q3	Practice 3.2 · Q3
NCERT Exercise 3.2 Q4	Practice 3.2 · Q4
NCERT Exercise 3.2 Q5	Practice 3.2 · Q5
NCERT Exercise 3.2 Q6	Practice 3.2 · Q6
NCERT Exercise 3.2 Q7	Practice 3.2 · Q7
NCERT Exercise 3.2 Q8	Practice 3.2 · Q8
NCERT Exercise 3.2 Q9	Practice 3.2 · Q9
NCERT Exercise 3.2 Q10	Practice 3.2 · Q10
NCERT Exercise 3.3 Q1	Practice 3.3 · Q1
NCERT Exercise 3.3 Q2	Practice 3.3 · Q2
NCERT Exercise 3.3 Q3	Practice 3.3 · Q3
NCERT Exercise 3.3 Q4	Practice 3.3 · Q4
NCERT Exercise 3.3 Q5	Practice 3.3 · Q5
NCERT Exercise 3.3 Q6	Practice 3.3 · Q6
NCERT Exercise 3.3 Q7	Practice 3.3 · Q7
NCERT Exercise 3.3 Q8	Practice 3.3 · Q8
NCERT Exercise 3.3 Q9	Practice 3.3 · Q9
NCERT Exercise 3.3 Q10	Practice 3.3 · Q10
NCERT Exercise 3.3 Q11	Practice 3.3 · Q11
NCERT Exercise 3.3 Q12	Practice 3.3 · Q12
NCERT Exercise 3.3 Q13	Practice 3.3 · Q13
NCERT Exercise 3.3 Q14	Practice 3.3 · Q14
NCERT Exercise 3.3 Q15	Practice 3.3 · Q15
NCERT Exercise 3.3 Q16	Practice 3.3 · Q16
NCERT Exercise 3.3 Q17	Practice 3.3 · Q17
NCERT Exercise 3.3 Q18	Practice 3.3 · Q18
NCERT Exercise 3.3 Q19	Practice 3.3 · Q19
NCERT Exercise 3.3 Q20	Practice 3.3 · Q20
NCERT Exercise 3.3 Q21	Practice 3.3 · Q21
NCERT Exercise 3.3 Q22	Practice 3.3 · Q22
NCERT Exercise 3.3 Q23	Practice 3.3 · Q23
NCERT Exercise 3.3 Q24	Practice 3.3 · Q24
NCERT Exercise 3.3 Q25	Practice 3.3 · Q25
NCERT Miscellaneous Exercise Q1	Chapter 3 par miscellaneous exercises · Q1
NCERT Miscellaneous Exercise Q2	Chapter 3 par miscellaneous exercises · Q2
NCERT Miscellaneous Exercise Q3	Chapter 3 par miscellaneous exercises · Q3
NCERT Miscellaneous Exercise Q4	Chapter 3 par miscellaneous exercises · Q4
NCERT Miscellaneous Exercise Q5	Chapter 3 par miscellaneous exercises · Q5
NCERT Miscellaneous Exercise Q6	Chapter 3 par miscellaneous exercises · Q6
NCERT Miscellaneous Exercise Q7	Chapter 3 par miscellaneous exercises · Q7
NCERT Miscellaneous Exercise Q8	Chapter 3 par miscellaneous exercises · Q8
NCERT Miscellaneous Exercise Q9	Chapter 3 par miscellaneous exercises · Q9
NCERT Miscellaneous Exercise Q10	Chapter 3 par miscellaneous exercises · Q10

Quick check: MCQs

3.1 Introduction2 questions
3.2.4 Degree aur radian ka relation4 questions
3.3 Trigonometric Functions3 questions
3.3.1 Trigonometric functions ke signs3 questions
3.3.2 Trigonometric functions ka domain aur range3 questions
Sum ka tangent aur cotangent4 questions
Double aur triple angles3 questions
Sums se products, products se sums3 questions
Miscellaneous examples3 questions

Glossary · 13

Key terms (English + Hindi)

allied angles: $\frac{π}{2} \pm x$ , $π \pm x$ aur $2 π - x$ jaise angles, jinki trigonometric values sign ke fark ke saath $x$ wali hoti hain, aur $\frac{π}{2} \pm x$ par sine aur cosine aapas mein badal jaate hain.Chapter mein pehli baar yahan aaya →
angleकोणkon: Turning ki woh amount jo ek ray ko vertex ke around uski initial side se terminal side tak le jaati hai; anticlockwise turns positive hote hain.Chapter mein pehli baar yahan aaya →
asymptoteअनंतस्पर्शीanantasparshi: Aisi line jiske graph paas aur paas aata jaata hai lekin use kabhi milta nahi, jaise $y = \tan x$ ke liye $x = \frac{π}{2}$ .Chapter mein pehli baar yahan aaya →

degreeअंशansh: Ek full turn ka 360vaan hissa, likhte hain $1^{\circ}$ ; $1^{\circ} = 60^{'}$ aur $1^{'} = 60^{''}$ .Chapter mein pehli baar yahan aaya →

even functionसम फलनsam phalan: Aisa function jiske liye har $x$ par $f (- x) = f (x)$ , jaise $\cos x$ .Chapter mein pehli baar yahan aaya →

odd functionविषम फलनvisham phalan: Aisa function jiske liye har $x$ par $f (- x) = - f (x)$ , jaise $\sin x$ aur $\tan x$ .Chapter mein pehli baar yahan aaya →

periodic functionआवर्ती फलनaavarti phalan: Aisa function jiske liye kisi fixed $p > 0$ ke saath saare $x$ par $f (x + p) = f (x)$ ; aisa sabse chhota $p$ period hai.Chapter mein pehli baar yahan aaya →

quadrantal angle: Aisa angle jo $\frac{π}{2}$ ka whole-number multiple ho, jisse uski terminal side kisi axis ke along ho.Chapter mein pehli baar yahan aaya →

radianरेडियनradian: Circle ke centre par woh angle jo radius jitni lambi arc banati hai; ek full turn $2 π$ radians hota hai.

sum-to-product formulas: Aise formulas jo do sines ya cosines ke sum ya difference ko product mein badalte hain, jaise $\cos x + \cos y = 2 \cos \frac{x + y}{2} \cos \frac{x - y}{2}$ .Chapter mein pehli baar yahan aaya →

trigonometric functionत्रिकोणमितीय फलनtrikonmitiya phalan: Real number $x$ ka $\sin x$ ya $\cos x$ jaisa function, jo unit circle par arc length $x$ wale point se padha jaata hai.
trigonometric identityत्रिकोणमितीय सर्वसमिकाtrikonmitiya sarvasamika: Trigonometric functions wali aisi equation jo variable ki har us value ke liye sahi hai jahan dono sides defined hain.Chapter mein pehli baar yahan aaya →

unit circleएकक वृत्तekak vritt: Radius 1 wala circle jiska centre origin par hai.Chapter mein pehli baar yahan aaya →

Trigonometric Functions: Class 11 Maths Chapter 3 Hinglish mein

Is chapter mein aap kya seekhenge

3.1 Introduction

3.2 Angles

3.2.1 Degree measure

3.2.2 Radian measure

3.2.3 Radian aur real numbers ka relation

3.2.4 Degree aur radian ka relation

Practice 3.1: answers ke saath

3.3 Trigonometric Functions

3.3.1 Trigonometric functions ke signs

3.3.2 Trigonometric functions ka domain aur range

Practice 3.2: answers ke saath

3.4 Do Angles ke Sum aur Difference ke Trigonometric Functions

Difference aur sum ka cosine

Sum ka sine, aur allied angles

Sum ka tangent aur cotangent

Double aur triple angles

Sums se products, products se sums

Practice 3.3: answers ke saath

Miscellaneous examples

Summary aur key points

Practice questions answers ke saath (NCERT “Chapter 3 par miscellaneous exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye