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MathematicsCh 3 Trigonometric Functions

Trigonometric Functions: Class 11 Maths · Chapter 3

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 100 min
  • 1 simulation
  • 59 practice questions
  • 28 MCQs

What you will learn in Trigonometric Functions

  • Measure angles as rotations in degrees and radians, convert between the two units, and use l = rθ for arcs.
  • Define the six trigonometric functions of any real number with the unit circle and evaluate them at quadrantal and standard angles.
  • Find the signs, domains, ranges and periods of the trigonometric functions and sketch their graphs.
  • Find all six values from one value and the quadrant, and evaluate the functions at large positive or negative angles.
  • Prove and apply the sum and difference formulas, the allied-angle results and the double- and triple-angle formulas.
  • Use the sum-to-product and product-to-sum formulas to simplify expressions and prove identities.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 3, pp. 43–75. Not affiliated with or endorsed by NCERT.

3.1 Introduction

At a Dussehra mela, keep your eye on one seat of the giant wheel. The wheel turns at a steady rate and the seat rises, falls and rises again, repeating the same pattern of heights on every turn. A right triangle cannot describe this. Once the seat passes the top, it has turned through more than 90∘, and after a few rounds through more than 1000∘.

5/5One full turn draws one complete wave; the next turn repeats it.

Fig. 3.1The height of a seat on a turning wheel traces a repeating wave.

Left: a giant wheel on a stand with one seat marked by a dot on a highlighted spoke. Right: axes with height up and angle turned along. Step 1: the seat is level with the hub and the graph point is at the origin. Step 2: after a quarter turn the seat is at the top and the first quarter of a wave rises to its peak. Step 3: after half a turn the seat is back at hub level and the wave returns to the axis. Step 4: after three quarters of a turn the seat is at the bottom and the wave reaches its lowest point. Step 5: after a full turn the seat is back where it started and one complete wave has been drawn.

Anything that goes round, or swings to and fro, makes a pattern like this. The 50-hertz alternating current in Indian homes, the tide rising and falling at Mumbai, a vibrating sitar string and a satellite circling the Earth are all described with the functions of this chapter. Their parent subject, trigonometry (त्रिकोणमिति), still carries its older job in its name: the Greek words behind it mean triangle and measure.

Your Class 10 tools cannot follow the seat for long. There, sin⁡θ was opposite over hypotenuse in a right triangle, which only makes sense while θ is acute. We keep everything that worked, the values at 30∘, 45∘ and 60∘ and the identity sin⁡2θ+cos⁡2θ=1, and remove the limit on the angle. The plan: measure angles as amounts of turning (in degrees and in a new unit, the radian), define sine and cosine for every real number with a circle, draw their graphs, and find the sine and cosine of a sum of two angles.

Quick check
1 of 2
A ladder 5 m long leans against a wall. Its foot is 3 m from the wall, and it makes an angle θ with the ground. What is cos⁡θ?
Show answer

Answer 0.6

cos⁡θ=adjacenthypotenuse=35=0.6. The wall height of 4 m is the opposite side, which gives sin⁡θ=0.8 instead.

3.2 Angles

Picture a door swinging on its hinge. In this chapter an angle is an amount of turning like that: a ray starts in one position, called its initial side, and turns about its endpoint, the vertex, until it reaches its terminal side. A sign records the direction, positive for anticlockwise turning and negative for clockwise. Nothing stops the ray from going round more than once, so an angle may be larger than a full turn (Fig. 3.2).

2/4An anticlockwise turn of 60∘ gives a positive angle.

Fig. 3.2An angle is an amount of turning, with a sign for its direction.NCERT Fig. 3.1 · p. 43

A fixed horizontal ray labelled initial side starts at a point labelled vertex O; a second, coloured ray turns about O. Step 1: the turning ray lies on the initial side. Step 2: it has turned 60 degrees anticlockwise, with a curved arrow marked plus 60 degrees and the note anticlockwise turn: positive. Step 3: it has turned 45 degrees clockwise instead, with an arrow marked minus 45 degrees and the note clockwise turn: negative. Step 4: a spiral arrow goes once round and on to 60 degrees, marked 420 degrees, with the note one full turn, then 60 degrees more.

Turning can be counted in whole turns, or revolutions; motors and fans are rated this way, in revolutions per minute, and a ceiling fan on full speed manages a few hundred. Whole turns are too coarse for most geometry, so two finer units are used instead: the degree and the radian.

3.2.1 Degree measure

One degree (अंश), written 1∘, is 1360 of a full turn. Finer still are the minute, a sixtieth of a degree, and the second, a sixtieth of a minute:

1∘=60′,1′=60′′

A full turn is therefore 360∘, half a turn 180∘ and a quarter turn 90∘. An angle of 450∘ is a full turn followed by a quarter turn, while −135∘ is three-eighths of a turn clockwise. Minutes and seconds still appear in astronomy and map-making; the Earth's axis, for example, is tilted at about 23∘26′ to the perpendicular of its orbit.

3.2.2 Radian measure

The number 360 is a human choice with no special meaning for a circle. A unit taken from the circle itself is more useful. Lay a string exactly as long as the radius along the rim (Fig. 3.3). The angle it covers at the centre is the same for every circle, large or small, and it is called one radian.

Definition
Radian

The angle at the centre of a circle cut off by an arc as long as the radius. On the unit circle (radius 1), an arc of length 1 marks 1 radian.

ExampleAn arc 5 cm long on a circle of radius 5 cm subtends exactly 1 radian at the centre.

2/4Laid along the rim, it marks an angle of 1 radian at the centre.

Fig. 3.3One radian is the angle cut off by an arc as long as the radius.NCERT Fig. 3.4 · p. 45

A circle with centre O and a radius OA labelled r. Step 1: a straight string of the same length r stands upright at A. Step 2: the string is laid along the rim from A, its end joined to O; the shaded sector has an angle marked 1, with the notes arc equals r and theta equals 1 radian. Step 3: a string twice as long covers an angle marked 2, with arc equals 2r and theta equals 2 radians. Step 4: tick marks at 1, 2, 3, 4, 5 and 6 radians run round the rim with a small gap left over, and the note says about 6.28 radii fit round the rim, so a full turn is 2 pi radians.

The circumference is 2πr, so exactly 2π radius-lengths fit round the rim, and one complete turn measures 2π radians. The angle at the centre grows in step with the arc it cuts off, so an arc of length l, which is l/r radius-lengths long, subtends l/r radians. This is the arc-length formula:

Arc length
l=rθ
(3.1)
where
l
length of the arc
r
radius of the circle
θ
angle at the centre, in radians

3.2.3 Relation between radian and real numbers

Radians let us treat angles as plain real numbers. Draw the unit circle with centre O and mark A=(1,0). Stand a number line at A so that it touches the circle, with 0 at A and the positive numbers going up (Fig. 3.4). Now wrap the positive half anticlockwise round the circle and the negative half clockwise, like thread round a spool.

4/4The number 2 marks an angle of 2 radians at O.

Fig. 3.4Wrapping the real line round the unit circle pairs every real number with an angle in radians.NCERT Fig. 3.5 · p. 46

A unit circle with centre O and a vertical number line touching it at A, where the line reads 0; 1 and 2 are above, minus 1 and minus 2 below. Step 1: only the straight number line is shown. Step 2: the upper half fades and its copy lies wrapped anticlockwise round the circle, with the points 1 and 2 marked on the rim and pi marked at the point opposite A. Step 3: the lower half is wrapped clockwise, with minus 1 and minus 2 marked below the axis. Step 4: radii join O to A and to the point 2, with the angle between them marked, and the note says the real number 2 marks an angle of 2 radians.

Each real number t lands on one point of the circle, and the arc from A to that point has length ∣t∣. So t marks an angle of t radians: positive numbers give anticlockwise angles and negative numbers clockwise ones. Different numbers can land on the same point, such as t and t+2π, but every real number is the radian measure of exactly one angle and every angle has a radian measure. That is why we can feed any real number into sine and cosine.

3.2.4 Relation between degree and radian

A full turn is 360∘ and also 2π radians. Halving both gives the one fact you need for every conversion:

Degrees and radians
π rad=180∘
(3.2)

Dividing, 1 rad =(180π)∘≈57.2958∘, which is about 57∘17′45′′, and 1∘=π180 rad, about 0.01745 rad. With the rough value π≈227 the first figure comes out as about 57∘16′, which is close enough for most problems. To convert, multiply by the right fraction:

Converting
y∘=πy180 radx rad=(180xπ)∘
(3.3)
Table 3.1Common angles in both units
DegreesRadiansFraction of a turn
30∘π6112
45∘π418
60∘π316
90∘π214
120∘2π313
135∘3π438
150∘5π6512
180∘π12
270∘3π234
360∘2π1
Example 1NCERT Example 1 · p. 47

Degrees and minutes to radians

Convert 47∘15′ into radians.

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 47

Radians to degrees, minutes and seconds

Convert 5 radians into degrees, minutes and seconds, taking π=227.

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 48

Radius from an arc

A slice of a round pizza has a tip angle of 45∘ and its crust is 11 cm long. What is the radius of the pizza? Take π=227.

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 48

The tip of a wiper blade

A windscreen wiper is 42 cm long from its pivot to its tip and sweeps through 120∘. How far does the tip travel in one sweep? Take π=227.

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 48

Same arc, two circles

Two round flower beds in a park are edged with identical 3 m strips of metal. One strip spans 40∘ at the centre of its bed, the other 72∘. Compare the radii of the two beds.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
Write 150∘ in radians as kπ and type k (a fraction such as 3/4 is fine).
π
Show answer

Answer 0.8333 π

150×π180=5π6, so k=56.
Practice · 9 questions

Practice 3.1: questions with answers

  1. Short answerEasyNCERT Exercise 3.1 Q1

    Convert into radians.

    1. 35∘

      Show solution

      Answer7π36

      35×π180=7π36
    2. −62∘30′

      Show solution

      Answer−25π72

      −6212×π180=−125π360
      =−25π72
    3. 225∘

      Show solution

      Answer5π4

      225×π180=5π4
    4. 600∘

      Show solution

      Answer10π3

      600×π180=10π3
  2. Short answerMediumNCERT Exercise 3.1 Q2

    Convert into degrees (and minutes and seconds where needed), taking π=227.

    1. 1124 rad

      Show solution

      Answer26∘15′

      1124×63011=2614∘

      A quarter of a degree is 15′.

    2. −3 rad

      Show solution

      Answerabout −171∘49′5′′

      −3×63011=−171911∘
      911×60′=49111′

      and 111×60′′≈5′′.

    3. 4π3

      Show solution

      Answer240∘

      4π3×180π=240∘
    4. 11π6

      Show solution

      Answer330∘

      11π6×180π=330∘
  3. NumericalEasyNCERT Exercise 3.1 Q3

    A grinder disc spins at 450 revolutions per minute. Find the angle, in radians, that it sweeps out each second.

    Show solution

    Answer15π rad, about 47.1 rad

    45060=7.5 rev/s
    7.5×2π=15π≈47.1 rad
  4. NumericalEasyNCERT Exercise 3.1 Q4

    A railway track curves along a circular arc of radius 1050 m. Through how many degrees does the direction of a train turn along 330 m of this track? Take π=227.

    °
    Show solution

    Answer18∘

    The direction of travel (the tangent) turns through the same angle as the radius, the angle at the centre.

    θ=3301050=1135 rad
    1135×63011=18∘
  5. NumericalMediumNCERT Exercise 3.1 Q5

    In a circle of radius 9 cm, a chord is 93 cm long. Find the length of the minor arc cut off by the chord.

    Show solution

    Answer6π cm, about 18.8 cm

    Half the chord is 932 and the radius is 9, so half the central angle α has

    sin⁡α=93/29=32

    Hence α=60∘ and the central angle is 120∘=2π3.

    l=9×2π3=6π cm
  6. Short answerEasyNCERT Exercise 3.1 Q6

    A jeweller bends two equal lengths of gold wire along two bangles of different sizes. One piece covers 50∘ of its bangle, measured at the centre, and the other covers 80∘. Find the ratio of the radii of the bangles.

    Show solution

    Answer8 : 5

    r1r2=θ2θ1=8050=85

    The radians conversion factor cancels, so degrees can be used directly in the ratio.

  7. NumericalEasyNCERT Exercise 3.1 Q7

    A playground swing hangs on chains 2.4 m long. Find the angle, in radians, through which the chains turn when the seat moves along an arc of the length given.

    1. 0.6 m

      rad
      Show answer

      Answer 0.25 rad

      θ=0.62.4=14 rad.
    2. 1.2 m

      rad
      Show answer

      Answer 0.5 rad

      θ=1.22.4=12 rad.
    3. 2 m

      rad
      Show answer

      Answer 0.8333 rad

      θ=22.4=56 rad.
  8. Quick answerMedium

    What is the angle between the hands of a clock at 4:20? Write it as kπ radians and type k.

    π
    Show answer

    Answer 0.0556 π

    The minute hand is at 120∘ from 12. The hour hand is at 4×30∘+20×0.5∘=130∘. The angle is 10∘=π18, so k=118.
  9. Quick answerMedium

    A wheel of radius 35 cm rolls 11 m along a road without slipping. Through how many radians does it turn?

    rad
    Show answer

    Answer 31.43 rad

    θ=lr=110035=2207≈31.4 rad, which is 5 full turns when π is taken as 227.

3.3 Trigonometric Functions

Place a right triangle so that its acute angle x sits at the centre of a circle of radius 1 and its hypotenuse is a radius. The hypotenuse is 1, so cos⁡x (adjacent over hypotenuse) equals the horizontal coordinate of the corner on the circle, and sin⁡x equals its vertical coordinate. That observation tells us how to define sine and cosine when there is no triangle at all: read them off the circle.

Let the unit circle have its centre O at the origin, and let A=(1,0). For any real number x, travel round the circle from A through an arc of length x, anticlockwise if x>0 and clockwise if x<0, and call the point you reach P(a,b). The angle AOP is then x radians (Fig. 3.5).

Definition
Sine and cosine of a real number

If the point at arc length x from A(1,0) on the unit circle is P(a,b), then cos⁡x=a and sin⁡x=b.

ExampleThe point at arc length π2 is (0,1), so cos⁡π2=0 and sin⁡π2=1.

3/4The coordinates of P are called cos⁡x and sin⁡x.

Fig. 3.5Sine and cosine are the coordinates of a point on the unit circle.NCERT Fig. 3.6 · p. 49

Axes with a circle of radius 1 centred at the origin O, and the point A at (1, 0). Step 1: a coloured arc of length x runs anticlockwise from A to a point P in the first quadrant. Step 2: a perpendicular from P meets the x-axis at M, shading the right triangle OMP with sides a along the axis, b upwards and hypotenuse 1; the panel reads a squared plus b squared equals 1. Step 3: the panel renames the coordinates: cos x equals a and sin x equals b, so cos squared x plus sin squared x equals 1. Step 4: the triangle fades and four dots mark A, B, C and D where the circle meets the axes, at angles 0, pi by 2, pi and 3 pi by 2.

Triangle OMP has a right angle at M, with OM=∣a∣, MP=∣b∣ and OP=1. Pythagoras gives a2+b2=1 for every point of the circle, in every quadrant, because squaring removes the signs. In the new notation:

Pythagorean identity
cos⁡2x+sin⁡2x=1
(3.4)

Here cos⁡2x means (cos⁡x)2. The identity holds for every real x, not just for acute angles.

The last step of Fig. 3.5 marks the four points where the circle meets the axes. They are reached by whole-number multiples of π2, which are therefore called quadrantal angles, and their sines and cosines can simply be read off. The pair (cos⁡x,sin⁡x) is (1,0) at x=0, (0,1) at x=π2, (−1,0) at x=π and (0,−1) at x=3π2; at x=2π we are back at (1,0).

An arc of length x+2π is an arc of length x followed by one extra lap, and a lap ends where it began. The same holds for any whole number of laps, forwards or backwards, so

Whole turns change nothing
sin⁡(x+2nπ)=sin⁡xcos⁡(x+2nπ)=cos⁡x
(3.5)
where
n
any integer

The sine is the height of P, so it is zero exactly when P is at A or C, that is, when x is a multiple of π. The cosine vanishes exactly when P is at B or D, that is, when x is π2 times an odd integer:

Zeros of sine and cosine
sin⁡x=0⇔x=nπcos⁡x=0⇔x=(2n+1)π2
(3.6)
where
n
any integer

The other four functions are built from sine and cosine. Each one is defined wherever its denominator is not zero:

The other four functions
tan⁡x=sin⁡x/cos⁡xcot⁡x=cos⁡x/sin⁡xsec⁡x=1/cos⁡xcosec⁡x=1/sin⁡x
(3.7)
where
tan⁡x, sec⁡x
defined when x≠(2n+1)π2
cot⁡x, cosec⁡x
defined when x≠nπ

Two more Pythagorean identities follow at once. Divide (3.4) by cos⁡2x, which is allowed whenever cos⁡x≠0; then divide it instead by sin⁡2x, allowed whenever sin⁡x≠0:

More Pythagorean identities
1+tan⁡2x=sec⁡2x1+cot⁡2x=cosec⁡2x
(3.8)

For 0<x<π2 the triangle OMP is exactly the right triangle of Class 10, so the new definitions agree with the old ratios and the familiar values carry over (Table 3.2). A handy way to remember the sine column: it reads 02,12,22,32,42, and the cosine column is the same list in reverse.

Table 3.2Values at the standard angles of the first quadrant
xsin⁡xcos⁡xtan⁡x
0010
π6123213
π412121
π332123
π210not defined

For the reciprocal functions, flip the entries: for instance sec⁡π3=2 and cot⁡π6=3, while cosec⁡0 does not exist.

Quick check
1 of 3
What is sin⁡3π2?
Show answer

Answer (D) −1

Three quarters of a turn ends at D(0,−1), and the sine is the second coordinate, −1. The answer 0 is cos⁡3π2, the first coordinate.

3.3.1 Sign of trigonometric functions

Compare the point P(a,b) at arc length x with the point Q at arc length −x. The two arcs are equally long but run in opposite directions from A, so Q is the mirror image of P in the x-axis: Q=(a,−b) (Fig. 3.6). This works in every quadrant, so for all real x:

Negative angles
cos⁡(−x)=cos⁡xsin⁡(−x)=−sin⁡x
(3.9)

3/3Same first coordinate, opposite second: cosine is even, sine is odd.

Fig. 3.6Reversing the direction of the arc reflects the point in the x-axis.NCERT Fig. 3.7 · p. 51

A unit circle with axes. Step 1: an anticlockwise arc labelled x runs from A to the point P(a, b) in the second quadrant. Step 2: a clockwise arc labelled minus x, of the same length, runs from A to the point Q(a, minus b) in the third quadrant, and a dashed vertical line joins P and Q across the x-axis. Step 3: the panel reads same first coordinate, opposite second, followed by cos of minus x equals cos x and sin of minus x equals minus sin x.

A function with f(−x)=f(x) for every x is an even function; one with f(−x)=−f(x) is an odd function. So cosine is even and sine is odd. Dividing the two rules shows that tan⁡(−x)=−tan⁡x, so tangent is odd as well.

Because P never leaves the unit circle, neither coordinate can exceed 1 in size: ∣sin⁡x∣≤1 and ∣cos⁡x∣≤1 for every x. The signs of the two coordinates depend only on the quadrant that P is in, and the other four functions, being built from sine and cosine, inherit their signs. Fig. 3.7 walks round the four quadrants.

4/4In IV, P=(+,−): only cosine and secant are positive.

Fig. 3.7The positive functions in quadrants I to IV spell A, S, T, C (remember Add Sugar To Coffee).NCERT p. 52

Axes divide the plane into four tiles, with a turning arm from the origin that points into the current quadrant. Step 1: quadrant I is highlighted and shows the letter A, the words all six and the sign pair plus, plus. Step 2: the arm turns into quadrant II, which shows S, sin and cosec, and minus, plus. Step 3: quadrant III shows T, tan and cot, and minus, minus. Step 4: quadrant IV shows C, cos and sec, and plus, minus; all four tiles are now labelled.

Table 3.3Signs of the six functions by quadrant
QuadrantPositiveNegative
Iall sixnone
IIsin, coseccos, sec, tan, cot
IIItan, cotsin, cosec, cos, sec
IVcos, secsin, cosec, tan, cot

The same reasoning works on a circle of any radius r. If P(a,b) lies on that circle and OP makes an angle x with the positive x-axis, then cos⁡x=ar and sin⁡x=br, because triangle OMP is the unit-circle triangle scaled up by r. The simulation uses r=5, where grid points such as (3,4) and (−4,3) give exact fractions.

SimulationPoints on a circle of radius 5

A coordinate grid from minus 6 to 6 on both axes with the four quadrants labelled. Point O is fixed at the origin and point P starts at (3, 4); the distance OP is drawn and measured, and the mirror image of P in the x-axis is shown. Dragging P snaps it to grid points and updates its coordinates and the distance.

Fig. 3.8Drag P to grid points with OP=5 and compare it with its mirror image in the x-axis.
Try this
  • Move P to (−4,3). Check that OP=5, then find cos⁡x=−45 and sin⁡x=35. Which quadrant is this?
  • Read the mirror image of P. Which of cos⁡x and sin⁡x changes sign, and which stays the same?
  • Put P at (0,−5), then at (−5,0). What are sin⁡x and cos⁡x at these quadrantal angles?
Quick check
1 of 3
If sin⁡x<0 and tan⁡x>0, in which quadrant does x end?
Show answer

Answer (C) III

Sine is negative in III and IV; tangent is positive in I and III. Only III fits. Quadrant IV is tempting, but there the tangent is negative.

3.3.2 Domain and range of trigonometric functions

Sine and cosine are defined for every real number, since every x gives a point on the circle, and their values are coordinates of that point, so both have range [−1,1]. The other four functions lose the points where their denominators vanish. Because the reciprocal of a number in [−1,1] lies outside (−1,1), secant and cosecant never take values strictly between −1 and 1. Table 3.4 collects the results; n stands for any integer.

Table 3.4Domains and ranges
FunctionDomainRange
sin⁡xℝ[−1,1]
cos⁡xℝ[−1,1]
tan⁡xℝ−{(2n+1)π2}ℝ
cot⁡xℝ−{nπ}ℝ
sec⁡xℝ−{(2n+1)π2}(−∞,−1]∪[1,∞)
cosec⁡xℝ−{nπ}(−∞,−1]∪[1,∞)

As x runs from 0 to 2π, the point P goes once round the circle, and you can read off how each function changes. The height sin⁡x climbs from 0 to 1 in quadrant I, falls back to 0 in II, drops to −1 in III and returns to 0 in IV. The cosine starts at 1, falls steadily to −1 at x=π and then climbs back to 1. The tangent increases in every quadrant: from 0 it grows without bound as x approaches π2,Writing tan⁡x→∞ is shorthand for this unbounded growth. ∞ is not a number, and tan⁡x never equals it. then returns from very large negative values just after π2 and rises again.

A function f is periodic if there is a number p>0 with f(x+p)=f(x) for every x in its domain; the smallest such p is its period. Sine and cosine have period 2π, and so do their reciprocals. Tangent and cotangent repeat sooner, after π: half a turn sends P(a,b) to (−a,−b), which leaves the ratio ba unchanged. We prove tan⁡(π+x)=tan⁡x in Section 3.4.

Plotting the values gives the graphs in Fig. 3.9, Fig. 3.10 and Fig. 3.11. Where a function is undefined, its graph runs alongside a vertical line, getting closer and closer without ever touching it. Such a line is called an asymptote.

2/4y=cos⁡x (dashed) has the same shape, with a peak at x=0.

Fig. 3.9Sine and cosine waves: same shape, shifted by a quarter turn.NCERT Figs 3.8, 3.9 · p. 54

Axes from minus pi to 3 pi with grid lines at heights 1 and minus 1. Step 1: the sine curve, a smooth wave through the origin that peaks at pi by 2 and dips to minus 1 at 3 pi by 2. Step 2: the cosine curve, dashed, is added; it has the same shape but peaks at 0 and at 2 pi. Step 3: an arrow from the cosine peak at 0 to the sine peak at pi by 2 is labelled pi by 2. Step 4: a double arrow under the axis from 0 to 2 pi is labelled one period equals 2 pi.

2/3y=cot⁡x (dashed) falls between asymptotes at the multiples of π.

Fig. 3.10Tangent and cotangent repeat every π and have vertical asymptotes.NCERT Figs 3.10, 3.11 · p. 54

Axes from minus pi to 2 pi, with heights up to about 3. Step 1: the tangent curve in three branches, each rising from far below to far above between dashed vertical asymptotes at minus pi by 2, pi by 2 and 3 pi by 2, crossing the axis at 0 and pi. Step 2: the tangent fades and the cotangent appears, dashed, falling in each branch between asymptotes at minus pi, 0, pi and 2 pi. Step 3: the tangent alone, with a double arrow from 0 to pi labelled period pi.

3/3Neither function takes values strictly between −1 and 1.

Fig. 3.11Secant and cosecant never take values strictly between −1 and 1.NCERT Figs 3.12, 3.13 · p. 55

Axes from minus pi to 2 pi. Step 1: the cosine curve is dotted, and the secant curve is drawn as U-shaped branches that touch the cosine curve at its peaks and troughs and run up or down beside dashed asymptotes at odd multiples of pi by 2. Step 2: the secant is replaced by the dotted sine curve and the cosecant branches, which touch the sine curve at its peaks and troughs, with asymptotes at multiples of pi. Step 3: a band between heights minus 1 and 1 is shaded and labelled no values here.

InteractiveUnit circle lab

A slider sets an angle from minus 720 to 720 degrees, with buttons to step by 15 degrees, to sweep through a full turn, and to show the angle in radians. A unit circle shows the point P, the arc from A to P (with a dashed ring and a note for extra full turns), the cosine as a solid blue segment along the x-axis and the sine as a dashed red segment upwards; in tangent mode a green segment on the tangent line at A shows tan x. Beside it a graph of sine, cosine or tangent from minus 4 pi to 4 pi marks the current angle. A panel names the quadrant and the signs of sin, cos and tan, gives exact values at multiples of 30 and 45 degrees, and says which angle between 0 and 360 degrees lands on the same point.

Fig. 3.12Turn the angle and watch P, its coordinates and the graph move together.

Use the lab to test the descriptions above. Press Play to sweep through a full turn and watch the dot trace the sine curve. Then switch the graph to tangent and watch the green segment on the tangent line shoot off as the angle approaches 90∘.

With the signs and the three Pythagorean identities, one value of a trigonometric function and the quadrant are enough to find all the others.

Example 6NCERT Example 6 · p. 55

Five values from a sine

Given sin⁡x=−817 with x in quadrant III, work out the remaining five trigonometric functions.

Try it yourself first. Then reveal the solution one step at a time.

Example 7NCERT Example 7 · p. 56

Five values from a tangent

Given tan⁡x=−2021 with x in quadrant II, work out the remaining five trigonometric functions.

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 56

A large angle in radians

Find cos⁡37π6.

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 57

A large negative angle in degrees

Find sin⁡(−1830∘).

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
Which of these is a possible value of sec⁡x?
Show answer

Answer (D) −32

∣sec⁡x∣=1∣cos⁡x∣≥1, so only −32 can occur. The values 12 and −23 are possible for cos⁡x, but not for its reciprocal.
Practice · 11 questions

Practice 3.2: questions with answers

  1. Short answerEasyNCERT Exercise 3.2 Q1

    Given cos⁡x=−3537 with x in quadrant III, work out the remaining five functions.

    Show solution

    Answersin⁡x=−1237, tan⁡x=1235, cot⁡x=3512, sec⁡x=−3735, cosec⁡x=−3712

    sin⁡2x=1−12251369=1441369

    In quadrant III the sine is negative, so sin⁡x=−1237.

    tan⁡x=−12−35=1235

    The other three are reciprocals.

  2. Short answerEasyNCERT Exercise 3.2 Q2

    Given sin⁡x=725 with x in quadrant II, work out the remaining five functions.

    Show solution

    Answercos⁡x=−2425, tan⁡x=−724, cot⁡x=−247, sec⁡x=−2524, cosec⁡x=257

    cos⁡2x=1−49625=576625

    In quadrant II the cosine is negative, so cos⁡x=−2425.

    tan⁡x=7−24=−724
  3. Short answerMediumNCERT Exercise 3.2 Q3

    Given cot⁡x=409 with x in quadrant III, work out the remaining five functions.

    Show solution

    Answertan⁡x=940, sin⁡x=−941, cos⁡x=−4041, cosec⁡x=−419, sec⁡x=−4140

    cosec⁡2x=1+160081=168181

    Sine is negative in quadrant III, so cosec⁡x=−419 and sin⁡x=−941.

    cos⁡x=cot⁡xsin⁡x=409(−941)=−4041
  4. Short answerEasyNCERT Exercise 3.2 Q4

    Given sec⁡x=6111 with x in quadrant IV, work out the remaining five functions.

    Show solution

    Answercos⁡x=1161, sin⁡x=−6061, tan⁡x=−6011, cot⁡x=−1160, cosec⁡x=−6160

    cos⁡x=1161
    sin⁡2x=1−1213721=36003721

    Sine is negative in quadrant IV, so sin⁡x=−6061 and tan⁡x=−6011.

  5. Short answerMediumNCERT Exercise 3.2 Q5

    Given tan⁡x=−4528 with x in quadrant II, work out the remaining five functions.

    Show solution

    Answersec⁡x=−5328, cos⁡x=−2853, sin⁡x=4553, cosec⁡x=5345, cot⁡x=−2845

    sec⁡2x=1+2025784=2809784

    In quadrant II the secant is negative: sec⁡x=−5328, so cos⁡x=−2853.

    sin⁡x=tan⁡xcos⁡x=4553
  6. Quick answerEasyNCERT Exercise 3.2 Q6

    Find cos⁡1125∘.

    Show answer

    Answer 0.7071

    1125∘−3×360∘=45∘, so the value is cos⁡45∘=12≈0.7071.
  7. Quick answerEasyNCERT Exercise 3.2 Q7

    Find sec⁡(−1380∘).

    Show answer

    Answer 2

    −1380∘+4×360∘=60∘, so the value is sec⁡60∘=2.
  8. Quick answerEasyNCERT Exercise 3.2 Q8

    Find tan⁡25π4.

    Show answer

    Answer 1

    25π4=6π+π4, so the value is tan⁡π4=1.
  9. Quick answerEasyNCERT Exercise 3.2 Q9

    Find cos⁡(−17π3).

    Show answer

    Answer 0.5

    −17π3+6π=π3, so the value is cos⁡π3=12.
  10. Quick answerEasyNCERT Exercise 3.2 Q10

    Find cosec⁡(−23π6).

    Show answer

    Answer 2

    −23π6+4π=π6, so the value is cosec⁡π6=2.
  11. Quick answerMedium

    Assertion (A): No real number x has sec⁡x=12. Reason (R): ∣cos⁡x∣≤1 for every real x.

    Show answer

    Answer (A) Both A and R are true, and R explains A.

    sec⁡x=12 would need cos⁡x=2, which R rules out. So A is true, R is true, and R is exactly the reason.

3.4 Trigonometric Functions of Sum and Difference of Two Angles

Knowing sin⁡x and sin⁡y does not give sin⁡(x+y) by addition: sin⁡30∘+sin⁡60∘≈1.37, yet sin⁡90∘=1. The correct rules, the sum and difference formulas, are the most useful trigonometric identities in the course. Two identities are already in hand from Section 3.3.1: sin⁡(−x)=−sin⁡x and cos⁡(−x)=cos⁡x. Everything else in this section grows from one more, the formula for cos⁡(x−y).

Cosine of a difference and of a sum

2/3Turning clockwise through y carries Q to A and P to R, at arc length x−y.

Fig. 3.13Turning a chord about the centre does not change its length.NCERT Fig. 3.14 · p. 58

A unit circle with centre O. Step 1: points P and Q on the circle at arc lengths x and y, labelled P(cos x, sin x) and Q(cos y, sin y), with radii to O, the chord PQ drawn thick, and the angle x minus y marked between the radii. Step 2: the triangle OPQ turns clockwise through the angle y; Q lands on A(1, 0) and P lands on R, the point at arc length x minus y, while a dotted copy of the old chord stays behind. Step 3: the note reads equal chords: same angle at O, and PQ equals RA.

On the unit circle take P(cos⁡x,sin⁡x) and Q(cos⁡y,sin⁡y), so that the radii OP and OQ make an angle x−y at O (Fig. 3.13). The distance formula, with cos⁡2+sin⁡2=1 used twice, gives

PQ2=(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2=2−2cos⁡xcos⁡y−2sin⁡xsin⁡y

Now turn triangle OPQ clockwise about O through the angle y. Q moves to A(1,0), and P moves to R, the point at arc length x−y, so R=(cos⁡(x−y),sin⁡(x−y)). Turning does not change lengths, so RA=PQ, and

RA2=(cos⁡(x−y)−1)2+sin⁡2(x−y)=2−2cos⁡(x−y)

Equating the two expressions and cancelling gives the result, valid for all real x and y:

Cosine of a difference
cos⁡(x−y) =cos⁡xcos⁡y+sin⁡xsin⁡y
(3.10)

Replacing y by −y, and using cos⁡(−y)=cos⁡y and sin⁡(−y)=−sin⁡y, gives the sum formula:

Cosine of a sum
cos⁡(x+y) =cos⁡xcos⁡y−sin⁡xsin⁡y
(3.11)

Sine of a sum, and allied angles

Put x=π2 in (3.10). Since cos⁡π2=0 and sin⁡π2=1, this gives cos⁡(π2−y)=sin⁡y. Replacing y by π2−y in that result gives sin⁡(π2−y)=cos⁡y. These are the co-function rules: the sine of an angle is the cosine of its complement.

Co-function rules
cos⁡(π2−x)=sin⁡xsin⁡(π2−x)=cos⁡x
(3.12)

The co-function rules turn a sine into a cosine, so the sine of a sum follows from (3.10). Open the steps to see how.

Sine of a sum or difference
sin⁡(x±y) =sin⁡xcos⁡y±cos⁡xsin⁡y
(3.13)

Choosing x or y to be a quadrantal angle in these formulas gives the values at π2+x, π±x and 2π−x in terms of those at x. Such angles are called allied angles. For example, sin⁡(π−x)=sin⁡πcos⁡x−cos⁡πsin⁡x=sin⁡x. The pictures in Fig. 3.14 make every case easy to recall, because each allied point is a reflection or a turn of P(a,b).

5/5A quarter turn gives the point at π2+x: (−b,a).

Fig. 3.14Allied angles: reflections and turns of P(a,b).NCERT p. 59

A unit circle with a point P(a, b) at a small angle x in the first quadrant. Step 1: only P, with sin x equals b and cos x equals a. Step 2: the y-axis is drawn as a dashed mirror and P is reflected to (minus a, b) at pi minus x, with sin of pi minus x equals sin x and cos of pi minus x equals minus cos x. Step 3: a dashed line through O joins P to (minus a, minus b) at pi plus x, a half turn. Step 4: the x-axis is the mirror and the image is (a, minus b) at 2 pi minus x. Step 5: a curved arrow shows a quarter turn to (minus b, a) at pi by 2 plus x, with sin of pi by 2 plus x equals cos x and cos of pi by 2 plus x equals minus sin x.

Table 3.5Allied angles in terms of x
Anglesincostan
π2−xcos⁡xsin⁡xcot⁡x
π2+xcos⁡x−sin⁡x−cot⁡x
π−xsin⁡x−cos⁡x−tan⁡x
π+x−sin⁡x−cos⁡xtan⁡x
2π−x−sin⁡xcos⁡x−tan⁡x

A quick test for signs: pretend x is a small acute angle, see which quadrant the allied angle is in, and give the answer the sign that the function has there.

Tangent and cotangent of a sum

Assume that none of the three cosines, of x, of y and of x+y, is zero. Write tan⁡(x+y) as a quotient of sine and cosine, expand both, and divide the top and bottom by cos⁡xcos⁡y. Open the steps for the details.

Tangent of a sum or difference
tan⁡(x±y) =tan⁡x±tan⁡y1∓tan⁡xtan⁡y
(3.14)

Cotangent works the same way. Provided the sines of x, of y and of x+y are all non-zero, divide by sin⁡xsin⁡y instead to get

Cotangent of a sum or difference
cot⁡(x±y) =cot⁡xcot⁡y∓1cot⁡y±cot⁡x
(3.15)
Example 10NCERT Example 10 · p. 64

Evaluating at standard angles

Show that 3tan⁡π4cosec⁡π6−4cos⁡2π3sec⁡π3=10.

Try it yourself first. Then reveal the solution one step at a time.

Example 11NCERT Example 11 · p. 64

An exact value from a sum

Find the exact value of cos⁡105∘.

Try it yourself first. Then reveal the solution one step at a time.

Example 12NCERT Example 12 · p. 64

A tangent beyond π

Find tan⁡19π12.

Try it yourself first. Then reveal the solution one step at a time.

Example 13NCERT Example 13 · p. 65

A quotient in tangents

Prove that cos⁡(x+y)cos⁡(x−y)=1−tan⁡xtan⁡y1+tan⁡xtan⁡y for all x and y at which both sides are defined.

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 65

A product of three tangents

Show that tan⁡5xtan⁡3xtan⁡2x=tan⁡5x−tan⁡3x−tan⁡2x wherever all three tangents are defined.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
sin⁡75∘ equals
Show answer

Answer (A) 6+24

sin⁡(45∘+30∘)=12⋅32+12⋅12=6+24≈0.966. Simply adding the sines gives about 1.21, more than 1, which no sine can be.

Double and triple angles

Setting y=x in the sum formulas gives the values at 2x. For the cosine, (3.4) lets us trade sin⁡2x and cos⁡2x for each other, so there are several equally useful forms.

Cosine of a double angle
cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x=1−tan⁡2x1+tan⁡2x
(3.16)

For the last form write cos⁡2x=cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x, since the denominator is 1, and divide the top and bottom by cos⁡2x; this needs cos⁡x≠0. The same trick turns sin⁡2x=2sin⁡xcos⁡x into a formula in tan⁡x.

Sine of a double angle
sin⁡2x=2sin⁡xcos⁡x=2tan⁡x1+tan⁡2x
(3.17)
Tangent of a double angle
tan⁡2x=2tan⁡x1−tan⁡2x
(3.18)

The tangent formula is (3.14) with y=x. It holds when tan⁡x is defined and 2x is not an odd multiple of π2, so that tan⁡2x exists too.

Two rearrangements of (3.16) are used again and again, often with x2 in place of x: cos⁡2x=1+cos⁡2x2 and sin⁡2x=1−cos⁡2x2. They turn squares into first powers, and they give half-angle values (Example 21).

Applying the sum formula once more, to 2x+x, gives the triple-angle formulas.

sin⁡3x=sin⁡(2x+x)=sin⁡2xcos⁡x+cos⁡2xsin⁡x=2sin⁡x(1−sin⁡2x)+sin⁡x−2sin⁡3x=3sin⁡x−4sin⁡3x

The third line uses sin⁡2x=2sin⁡xcos⁡x, cos⁡2x=1−sin⁡2x and cos⁡2x=1−2sin⁡2x. Expanding cos⁡(2x+x) and tan⁡(2x+x) in the same way gives the other two formulas; the tangent one needs 3x not to be an odd multiple of π2.

Triple angles
sin⁡3x=3sin⁡x−4sin⁡3xcos⁡3x=4cos⁡3x−3cos⁡xtan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x
(3.19)
Quick check
1 of 3
If cos⁡x=13, find cos⁡2x.
Show answer

Answer -0.7778

cos⁡2x=2cos⁡2x−1=29−1=−79≈−0.778. Doubling the cosine, 23, is the tempting wrong answer.

Sums into products, products into sums

Add the expansions of cos⁡(x+y) and cos⁡(x−y) and the sine terms cancel; subtract them and the cosine terms cancel. The same happens with sin⁡(x+y) and sin⁡(x−y). The four results are the product-to-sum formulas:

Product to sum
2cos⁡xcos⁡y =cos⁡(x+y)+cos⁡(x−y)−2sin⁡xsin⁡y =cos⁡(x+y)−cos⁡(x−y)2sin⁡xcos⁡y =sin⁡(x+y)+sin⁡(x−y)2cos⁡xsin⁡y =sin⁡(x+y)−sin⁡(x−y)
(3.20)

Now read them backwards. Put x+y=C and x−y=D, so that x=C+D2 and y=C−D2. Each formula then turns a sum or difference into a product. Renaming C and D as x and y gives the sum-to-product formulas:

Sum to product
cos⁡x+cos⁡y =2cos⁡x+y2cos⁡x−y2cos⁡x−cos⁡y =−2sin⁡x+y2sin⁡x−y2sin⁡x+sin⁡y =2sin⁡x+y2cos⁡x−y2sin⁡x−sin⁡y =2cos⁡x+y2sin⁡x−y2
(3.21)

In words: the sum of two cosines is twice the cosine of the half-sum times the cosine of the half-difference, and so on. Their main use is in proofs, wherever a sum has to be factorised.

They also explain a sound you may have heard. Two tuning forks of slightly different frequencies, struck together, give a note that swells and fades regularly; these swells are called beats. Adding two cosine waves of nearly equal frequency gives a fast wave whose size is controlled by a slow cosine factor (Fig. 3.15).

3/3The sum is 2cos⁡6xcos⁡x: the slow factor 2cos⁡x (dashed) is the envelope.

Fig. 3.15Adding two close frequencies produces beats.

Top panel: two thin waves, cos 7x drawn solid and cos 5x dashed, over one interval from 0 to 2 pi; they start together, drift out of step and come back. Step 2: a bottom panel adds their sum, a fast wave whose height swells to 2 where the top waves agree and shrinks to nearly 0 where they disagree. Step 3: two dashed curves, plus and minus 2 cos x, hug the sum from above and below, labelled 2 cos 6x cos x.

Example 15NCERT Example 15 · p. 65

A sum of two sines

Prove that sin⁡(π3+x)+sin⁡(π3−x)=3cos⁡x.

Try it yourself first. Then reveal the solution one step at a time.

Example 16NCERT Example 16 · p. 66

A quotient of sums

Prove that sin⁡9x+sin⁡3xcos⁡9x−cos⁡3x=−cot⁡3x.

Try it yourself first. Then reveal the solution one step at a time.

Example 17NCERT Example 17 · p. 66

Grouping before factorising

Prove that sin⁡3x+sin⁡x−2sin⁡2xcos⁡3x−cos⁡x=tan⁡x2.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
cos⁡75∘+cos⁡15∘ equals
Show answer

Answer (B) 32

2cos⁡45∘cos⁡30∘=2⋅12⋅32=32≈1.22. Adding the angles first, cos⁡90∘=0, is not allowed.
Practice · 25 questions

Practice 3.3: questions with answers

  1. ProofEasyNCERT Exercise 3.3 Q1

    Prove that cos⁡2π6+sin⁡2π4−cot⁡2π4=14.

    Show solution

    Here cos⁡2π6=34, sin⁡2π4=12 and cot⁡2π4=1.

    34+12−1=14
  2. ProofEasyNCERT Exercise 3.3 Q2

    Prove that 2cos⁡2π4+sec⁡25π6sin⁡2π6=43.

    Show solution

    cos⁡5π6=−32, so sec⁡25π6=43.

    2⋅12+43⋅14
    =1+13=43
  3. ProofEasyNCERT Exercise 3.3 Q3

    Prove that tan⁡2π3+cosec⁡7π6+3cot⁡2π3=2.

    Show solution

    sin⁡7π6=−sin⁡π6=−12, so cosec⁡7π6=−2.

    3−2+3⋅13=2
  4. ProofEasyNCERT Exercise 3.3 Q4

    Prove that 2cos⁡23π4+tan⁡22π3+2cosec⁡2π4=8.

    Show solution

    cos⁡3π4=−12, tan⁡2π3=−3, cosec⁡π4=2.

    2⋅12+3+2⋅2=8
  5. Short answerMediumNCERT Exercise 3.3 Q5

    Find the exact value of

    1. sin⁡105∘

      Show solution

      Answer6+24

      sin⁡(60∘+45∘)
      =32⋅12+12⋅12
      =6+24
    2. tan⁡165∘

      Show solution

      Answer3−2

      tan⁡165∘=tan⁡(180∘−15∘)=−tan⁡15∘, and

      tan⁡15∘=1−131+13
      =3−13+1=2−3

      so tan⁡165∘=3−2.

  6. ProofMediumNCERT Exercise 3.3 Q6

    Prove that cos⁡(π3+x)cos⁡(π6+y)−sin⁡(π3+x)sin⁡(π6+y)=−sin⁡(x+y).

    Show solution

    By (3.11) the left side is the cosine of the sum of the two angles:

    cos⁡(π2+x+y)

    and cos⁡(π2+θ)=−sin⁡θ, which gives −sin⁡(x+y).

  7. ProofMediumNCERT Exercise 3.3 Q7

    Prove that tan⁡(π4+x)−tan⁡(π4−x)=2tan⁡2x.

    Show solution

    With t=tan⁡x, (3.14) gives

    1+t1−t−1−t1+t
    =(1+t)2−(1−t)21−t2
    =4t1−t2=2tan⁡2x
  8. ProofMediumNCERT Exercise 3.3 Q8

    Prove that cos⁡(π2+x)sec⁡(−x)sin⁡(π−x)cot⁡(π2−x)=−cosec⁡x.

    Show solution

    The top is (−sin⁡x)sec⁡x=−tan⁡x, and the bottom is sin⁡xtan⁡x, so

    −tan⁡xsin⁡xtan⁡x=−1sin⁡x

    which is −cosec⁡x.

  9. ProofHardNCERT Exercise 3.3 Q9

    Prove that sin⁡(3π2−x)sin⁡(2π−x)[tan⁡(3π2+x)−tan⁡(π+x)]=−1.

    Show solution

    sin⁡(3π2−x)=−cos⁡x and sin⁡(2π−x)=−sin⁡x; their product is sin⁡xcos⁡x.

    tan⁡(3π2+x)=−cos⁡xsin⁡x=−cot⁡x and tan⁡(π+x)=tan⁡x, so the bracket is

    −cos⁡2x+sin⁡2xsin⁡xcos⁡x
    =−1sin⁡xcos⁡x

    Multiplying by sin⁡xcos⁡x gives −1.

  10. ProofEasyNCERT Exercise 3.3 Q10

    Prove that sin⁡(n+3)xcos⁡(n+1)x−cos⁡(n+3)xsin⁡(n+1)x=sin⁡2x.

    Show solution

    This is sin⁡(A−B) with A=(n+3)x and B=(n+1)x, and A−B=2x.

  11. ProofMediumNCERT Exercise 3.3 Q11

    Prove that cos⁡(2π3+x)−cos⁡(2π3−x)=−3sin⁡x.

    Show solution

    The half-sum is 2π3 and the half-difference is x:

    −2sin⁡2π3sin⁡x
    =−2⋅32sin⁡x=−3sin⁡x
  12. ProofMediumNCERT Exercise 3.3 Q12

    Prove that sin⁡25x−sin⁡22x=sin⁡7xsin⁡3x.

    Show solution

    Use sin⁡2θ=1−cos⁡2θ2:

    1−cos⁡10x2−1−cos⁡4x2
    =12(cos⁡4x−cos⁡10x)

    By (3.21), cos⁡4x−cos⁡10x=2sin⁡7xsin⁡3x, so the left side is sin⁡7xsin⁡3x.

  13. ProofMediumNCERT Exercise 3.3 Q13

    Prove that cos⁡2x−sin⁡23x=cos⁡4xcos⁡2x.

    Show solution

    Use cos⁡2x=1+cos⁡2x2 and sin⁡23x=1−cos⁡6x2:

    LHS=12(cos⁡6x+cos⁡2x)
    =12⋅2cos⁡4xcos⁡2x
    =cos⁡4xcos⁡2x
  14. ProofMediumNCERT Exercise 3.3 Q14

    Prove that sin⁡x+2sin⁡3x+sin⁡5x=4sin⁡3xcos⁡2x.

    Show solution
    sin⁡5x+sin⁡x=2sin⁡3xcos⁡2x
    LHS=2sin⁡3x(cos⁡2x+1)

    and 1+cos⁡2x=2cos⁡2x, so the left side is 4sin⁡3xcos⁡2x.

  15. ProofMediumNCERT Exercise 3.3 Q15

    Prove that tan⁡3x(cos⁡5x+cos⁡x)=sin⁡5x+sin⁡x.

    Show solution
    cos⁡5x+cos⁡x=2cos⁡3xcos⁡2x
    sin⁡5x+sin⁡x=2sin⁡3xcos⁡2x

    So tan⁡3x⋅2cos⁡3xcos⁡2x=2sin⁡3xcos⁡2x, the right side.

  16. ProofMediumNCERT Exercise 3.3 Q16

    Prove that cos⁡11x−cos⁡3xsin⁡15x−sin⁡x=−sin⁡4xcos⁡8x.

    Show solution
    cos⁡11x−cos⁡3x=−2sin⁡7xsin⁡4x
    sin⁡15x−sin⁡x=2cos⁡8xsin⁡7x

    Cancelling 2sin⁡7x leaves −sin⁡4xcos⁡8x.

  17. ProofEasyNCERT Exercise 3.3 Q17

    Prove that sin⁡7x−sin⁡xcos⁡7x+cos⁡x=tan⁡3x.

    Show solution
    2cos⁡4xsin⁡3x2cos⁡4xcos⁡3x=tan⁡3x
  18. ProofMediumNCERT Exercise 3.3 Q18

    Prove that cos⁡x−cos⁡ysin⁡x+sin⁡y=tan⁡y−x2.

    Show solution
    −2sin⁡x+y2sin⁡x−y22sin⁡x+y2cos⁡x−y2
    =−tan⁡x−y2

    and −tan⁡x−y2=tan⁡y−x2 because tangent is odd.

  19. ProofEasyNCERT Exercise 3.3 Q19

    Prove that sin⁡2x+sin⁡4xcos⁡2x−cos⁡4x=cot⁡x.

    Show solution
    2sin⁡3xcos⁡x−2sin⁡3xsin⁡(−x)

    Since sin⁡(−x)=−sin⁡x, this is cos⁡xsin⁡x=cot⁡x.

  20. ProofEasyNCERT Exercise 3.3 Q20

    Prove that cos⁡3x+cos⁡xcos⁡2x−sin⁡2x=2cos⁡x.

    Show solution
    2cos⁡2xcos⁡xcos⁡2x=2cos⁡x

    using (3.21) on top and (3.16) below.

  21. ProofMediumNCERT Exercise 3.3 Q21

    Prove that sin⁡x+sin⁡3x+sin⁡5xcos⁡x+cos⁡3x+cos⁡5x=tan⁡3x.

    Show solution
    sin⁡5x+sin⁡x=2sin⁡3xcos⁡2x
    cos⁡5x+cos⁡x=2cos⁡3xcos⁡2x
    sin⁡3x(2cos⁡2x+1)cos⁡3x(2cos⁡2x+1)

    which is tan⁡3x.

  22. ProofHardNCERT Exercise 3.3 Q22

    Prove that cot⁡xcot⁡3x−cot⁡3xcot⁡4x−cot⁡4xcot⁡x=1.

    Show solution
    cot⁡4x=cot⁡3xcot⁡x−1cot⁡x+cot⁡3x

    Cross-multiplying:

    cot⁡4xcot⁡x+cot⁡4xcot⁡3x=cot⁡3xcot⁡x−1

    Rearranging gives the result.

  23. ProofHardNCERT Exercise 3.3 Q23

    Prove that cos⁡4x=1−6tan⁡2x+tan⁡4x(1+tan⁡2x)2.

    Show solution

    With t=tan⁡x and tan⁡2x=2t1−t2:

    cos⁡4x=1−tan⁡22x1+tan⁡22x
    =(1−t2)2−4t2(1−t2)2+4t2

    The top is 1−6t2+t4 and the bottom is 1+2t2+t4=(1+t2)2.

  24. ProofMediumNCERT Exercise 3.3 Q24

    Prove that sin⁡4x+cos⁡4x=3+cos⁡4x4.

    Show solution
    (sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x
    =1−12sin⁡22x
    =1−14(1−cos⁡4x)
    =3+cos⁡4x4
  25. ProofHardNCERT Exercise 3.3 Q25

    Prove that cos⁡5x=16cos⁡5x−20cos⁡3x+5cos⁡x.

    Show solution

    Let c=cos⁡x. Then cos⁡3x=4c3−3c and cos⁡2x=2c2−1, so

    cos⁡5x=2cos⁡3xcos⁡2x−c
    =2(4c3−3c)(2c2−1)−c
    =16c5−20c3+5c

Miscellaneous examples

These examples combine several ideas from the chapter: quadrant signs, the sum formulas and the multiple-angle formulas. In each one, decide the quadrant before taking a square root.

Example 18NCERT Example 18 · p. 68

Combining two angles

If cos⁡x=−1517 with x in the second quadrant, and sin⁡y=−35 with y in the fourth quadrant, find cos⁡(x+y).

Try it yourself first. Then reveal the solution one step at a time.

Example 19NCERT Example 19 · p. 68

Products into sums and back

Prove that sin⁡x2sin⁡7x2+sin⁡3x2sin⁡11x2=sin⁡2xsin⁡5x.

Try it yourself first. Then reveal the solution one step at a time.

Example 20NCERT Example 20 · p. 69

An exact tangent from a quadratic

Find the exact value of tan⁡3π8.

Try it yourself first. Then reveal the solution one step at a time.

For half-angle questions, first locate x2: halving the inequality that places x in its quadrant tells you where x2 lies, and so which signs to choose (Fig. 3.16).

3/3So sin⁡x2>0 and cos⁡x2<0.

Fig. 3.16Halving the angle moves it from quadrant III to quadrant II.NCERT p. 70

A unit circle with axes and quadrant labels. Step 1: quadrant III is shaded and an arc from A runs anticlockwise to a point at angle x between pi and 3 pi by 2. Step 2: that arc fades; the sector from pi by 2 to 3 pi by 4 is shaded and a shorter arc runs to the point at angle x by 2 in quadrant II. Step 3: the panel reads x by 2 lies in II, so the sine of x by 2 is positive and its cosine is negative.

Example 21NCERT Example 21 · p. 70

Half-angle values

Given sin⁡x=−2425 with π<x<3π2, work out cos⁡x2, sin⁡x2 and tan⁡x2.

Try it yourself first. Then reveal the solution one step at a time.

Example 22NCERT Example 22 · p. 71

Squares of three sines

Prove that sin⁡2x+sin⁡2(x+2π3)+sin⁡2(x−2π3)=32.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
If sin⁡A=817 and sin⁡B=35 with A and B acute, find sin⁡(A+B) as a decimal.
Show answer

Answer 0.9059

cos⁡A=1517 and cos⁡B=45, so sin⁡(A+B)=817⋅45+1517⋅35=7785≈0.906.
Timeline

Milestones

  1. c. 140 BCE

    Hipparchus

    A table of chords for astronomy.

  2. 499 CE

    Aryabhata

    The Aryabhatiya gives a table for the half-chord (sine).

  3. c. 629

    Bhaskara I

    A simple fraction that approximates the sine closely.

  4. 1150

    Bhaskara II

    Siddhānta Śiromaṇi, with sines of sums and exact values.

  5. c. 1530

    Yuktibhāṣā

    Proofs and sine series from the Kerala school, in Malayalam.

  6. 1813

    John Herschel

    Introduces the notation sin⁡−1x for the inverse sine.

Recap

Trigonometric Functions: key points and summary

  1. An angle is an amount of turning: anticlockwise is positive, clockwise negative (Section 3.2).
  2. A degree is 1360 of a turn, with 1∘=60′ and 1′=60′′ (Section 3.2.1).
  3. An arc equal to the radius subtends 1 radian, and l=rθ with θ in radians (Section 3.2.2).
  4. π rad =180∘: multiply by π180 or by 180π to convert (Section 3.2.4).
  5. If the point at arc length x on the unit circle is (a,b), then cos⁡x=a, sin⁡x=b and cos⁡2x+sin⁡2x=1 (Section 3.3).
  6. Dividing by cos⁡2x or sin⁡2x gives the two other Pythagorean identities ((3.8)).
  7. Sine vanishes only at multiples of π, cosine only at odd multiples of π2 ((3.6)).
  8. Cosine is even; sine and tangent are odd. The positive functions by quadrant spell A, S, T, C (Section 3.3.1).
  9. Sine, cosine, secant and cosecant have period 2π; tangent and cotangent have period π (Section 3.3.2).
  10. The cosine of a difference comes from one chord of the unit circle ((3.10)); the cosine and sine of sums and differences follow ((3.11), (3.13)).
  11. Allied angles π2±x, π±x, 2π−x give the values at x up to sign, with sine and cosine swapping at π2±x (Sine of a sum, and allied angles).
  12. Dividing the sine formula by the cosine formula gives the tangent rule ((3.14)); the cotangent rule is its mirror ((3.15)).
  13. Putting y=x gives the double angles ((3.16), (3.17), (3.18)); one more step gives the triple angles ((3.19)).
  14. Adding and subtracting the sum formulas turns products into sums ((3.20)); reading them backwards factorises sums ((3.21)).
Practice · 14 questions

Practice questions with answers (NCERT “Miscellaneous exercises on Chapter 3”)

  1. ProofMediumNCERT Miscellaneous Exercise Q1

    Prove that 2cos⁡π7cos⁡4π7+cos⁡2π7+cos⁡4π7=0.

    Show solution
    2cos⁡π7cos⁡4π7=cos⁡5π7+cos⁡3π7

    Now cos⁡5π7=−cos⁡2π7 and cos⁡3π7=−cos⁡4π7, so all four terms cancel in pairs.

  2. ProofEasyNCERT Miscellaneous Exercise Q2

    Prove that (cos⁡4x+cos⁡2x)sin⁡x−(sin⁡4x−sin⁡2x)cos⁡x=0.

    Show solution
    cos⁡4x+cos⁡2x=2cos⁡3xcos⁡x
    sin⁡4x−sin⁡2x=2cos⁡3xsin⁡x

    Both terms equal 2cos⁡3xsin⁡xcos⁡x, so their difference is 0.

  3. ProofMediumNCERT Miscellaneous Exercise Q3

    Prove that (cos⁡x−cos⁡y)2+(sin⁡x+sin⁡y)2=4sin⁡2x+y2.

    Show solution
    2−2cos⁡xcos⁡y+2sin⁡xsin⁡y
    =2−2cos⁡(x+y)
    =4sin⁡2x+y2

    using 1−cos⁡θ=2sin⁡2θ2.

  4. ProofMediumNCERT Miscellaneous Exercise Q4

    Prove that (cos⁡x+cos⁡y)2+(sin⁡x+sin⁡y)2−(cos⁡x−cos⁡y)2−(sin⁡x−sin⁡y)2=4cos⁡(x−y).

    Show solution

    With p,q the two cosines, then the two sines:

    4cos⁡xcos⁡y+4sin⁡xsin⁡y
    =4cos⁡(x−y)

    by (3.10).

  5. ProofMediumNCERT Miscellaneous Exercise Q5

    Prove that cos⁡x+cos⁡3x+cos⁡5x+cos⁡7x=4cos⁡xcos⁡2xcos⁡4x.

    Show solution
    2cos⁡4xcos⁡3x+2cos⁡4xcos⁡x
    =2cos⁡4x(cos⁡3x+cos⁡x)
    =2cos⁡4x⋅2cos⁡2xcos⁡x
  6. ProofMediumNCERT Miscellaneous Exercise Q6

    Prove that (sin⁡8x+sin⁡2x)+(sin⁡6x+sin⁡4x)(cos⁡8x+cos⁡2x)+(cos⁡6x+cos⁡4x)=tan⁡5x.

    Show solution

    Top: 2sin⁡5xcos⁡3x+2sin⁡5xcos⁡x. Bottom: 2cos⁡5xcos⁡3x+2cos⁡5xcos⁡x.

    2sin⁡5x(cos⁡3x+cos⁡x)2cos⁡5x(cos⁡3x+cos⁡x)

    which is tan⁡5x.

  7. ProofHardNCERT Miscellaneous Exercise Q7

    Prove that sin⁡x+sin⁡2x−sin⁡3x=4sin⁡xsin⁡x2sin⁡3x2.

    Show solution
    sin⁡x+sin⁡2x=2sin⁡3x2cos⁡x2
    LHS=2sin⁡3x2(cos⁡x2−cos⁡3x2)

    and cos⁡x2−cos⁡3x2=2sin⁡xsin⁡x2, which gives the right side.

  8. Short answerMediumNCERT Miscellaneous Exercise Q8

    For x in quadrant II with tan⁡x=−158, work out cos⁡x2, sin⁡x2 and tan⁡x2.

    Show solution

    Answercos⁡x2=334, sin⁡x2=534, tan⁡x2=53

    sec⁡2x=1+22564=28964; in quadrant II, cos⁡x=−817.

    π4<x2<π2, so both half-angle values are positive.

    sin⁡2x2=1+8/172=2534
    cos⁡2x2=1−8/172=934

    so tan⁡x2=53.

  9. Short answerMediumNCERT Miscellaneous Exercise Q9

    For x in quadrant III with cos⁡x=−79, work out cos⁡x2, sin⁡x2 and tan⁡x2.

    Show solution

    Answercos⁡x2=−13, sin⁡x2=223, tan⁡x2=−22

    π2<x2<3π4: sine positive, cosine negative.

    sin⁡2x2=1+7/92=89
    cos⁡2x2=1−7/92=19

    so sin⁡x2=223, cos⁡x2=−13 and tan⁡x2=−22.

  10. Short answerHardNCERT Miscellaneous Exercise Q10

    For x in quadrant II with sin⁡x=13, work out cos⁡x2, sin⁡x2 and tan⁡x2.

    Show solution

    Answercos⁡x2=2−16, sin⁡x2=2+16, tan⁡x2=3+22

    In quadrant II, cos⁡x=−223, and x2 lies in quadrant I.

    sin⁡2x2=3+226=(2+1)26
    cos⁡2x2=3−226=(2−1)26

    Dividing the roots, tan⁡x2=2+12−1=3+22.

  11. Quick answerMedium

    The greatest value of 3sin⁡x+4cos⁡x is

    Show answer

    Answer (B) 5

    With cos⁡α=35 and sin⁡α=45, the expression is 5(sin⁡xcos⁡α+cos⁡xsin⁡α)=5sin⁡(x+α), whose greatest value is 5. Adding 3+4=7 assumes both terms peak at the same x, which never happens.
  12. Quick answerHard

    cos⁡20∘cos⁡40∘cos⁡80∘ equals

    Show answer

    Answer (B) 18

    8sin⁡20∘× the product =4sin⁡40∘cos⁡40∘cos⁡80∘=2sin⁡80∘cos⁡80∘=sin⁡160∘=sin⁡20∘. So the product is 18.
  13. Quick answerHard

    Assertion (A): The equation sin⁡x+cos⁡x=1.5 has no real solution. Reason (R): sin⁡x+cos⁡x=2sin⁡(x+π4).

    Show answer

    Answer (A) Both A and R are true, and R explains A.

    R is the sine-of-a-sum formula with cos⁡π4=sin⁡π4=12. It shows sin⁡x+cos⁡x≤2≈1.414<1.5, which is exactly why A is true.
  14. Quick answerMedium

    Find tan⁡15∘+cot⁡15∘.

    Show answer

    Answer 4

    sin⁡2+cos⁡2sin⁡15∘cos⁡15∘=2sin⁡30∘=4. Check: (2−3)+(2+3)=4.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercise 3.1 Q1Practice 3.1 · Q1
NCERT Exercise 3.1 Q2Practice 3.1 · Q2
NCERT Exercise 3.1 Q3Practice 3.1 · Q3
NCERT Exercise 3.1 Q4Practice 3.1 · Q4
NCERT Exercise 3.1 Q5Practice 3.1 · Q5
NCERT Exercise 3.1 Q6Practice 3.1 · Q6
NCERT Exercise 3.1 Q7Practice 3.1 · Q7
NCERT Exercise 3.2 Q1Practice 3.2 · Q1
NCERT Exercise 3.2 Q2Practice 3.2 · Q2
NCERT Exercise 3.2 Q3Practice 3.2 · Q3
NCERT Exercise 3.2 Q4Practice 3.2 · Q4
NCERT Exercise 3.2 Q5Practice 3.2 · Q5
NCERT Exercise 3.2 Q6Practice 3.2 · Q6
NCERT Exercise 3.2 Q7Practice 3.2 · Q7
NCERT Exercise 3.2 Q8Practice 3.2 · Q8
NCERT Exercise 3.2 Q9Practice 3.2 · Q9
NCERT Exercise 3.2 Q10Practice 3.2 · Q10
NCERT Exercise 3.3 Q1Practice 3.3 · Q1
NCERT Exercise 3.3 Q2Practice 3.3 · Q2
NCERT Exercise 3.3 Q3Practice 3.3 · Q3
NCERT Exercise 3.3 Q4Practice 3.3 · Q4
NCERT Exercise 3.3 Q5Practice 3.3 · Q5
NCERT Exercise 3.3 Q6Practice 3.3 · Q6
NCERT Exercise 3.3 Q7Practice 3.3 · Q7
NCERT Exercise 3.3 Q8Practice 3.3 · Q8
NCERT Exercise 3.3 Q9Practice 3.3 · Q9
NCERT Exercise 3.3 Q10Practice 3.3 · Q10
NCERT Exercise 3.3 Q11Practice 3.3 · Q11
NCERT Exercise 3.3 Q12Practice 3.3 · Q12
NCERT Exercise 3.3 Q13Practice 3.3 · Q13
NCERT Exercise 3.3 Q14Practice 3.3 · Q14
NCERT Exercise 3.3 Q15Practice 3.3 · Q15
NCERT Exercise 3.3 Q16Practice 3.3 · Q16
NCERT Exercise 3.3 Q17Practice 3.3 · Q17
NCERT Exercise 3.3 Q18Practice 3.3 · Q18
NCERT Exercise 3.3 Q19Practice 3.3 · Q19
NCERT Exercise 3.3 Q20Practice 3.3 · Q20
NCERT Exercise 3.3 Q21Practice 3.3 · Q21
NCERT Exercise 3.3 Q22Practice 3.3 · Q22
NCERT Exercise 3.3 Q23Practice 3.3 · Q23
NCERT Exercise 3.3 Q24Practice 3.3 · Q24
NCERT Exercise 3.3 Q25Practice 3.3 · Q25
NCERT Miscellaneous Exercise Q1Miscellaneous exercises on Chapter 3 · Q1
NCERT Miscellaneous Exercise Q2Miscellaneous exercises on Chapter 3 · Q2
NCERT Miscellaneous Exercise Q3Miscellaneous exercises on Chapter 3 · Q3
NCERT Miscellaneous Exercise Q4Miscellaneous exercises on Chapter 3 · Q4
NCERT Miscellaneous Exercise Q5Miscellaneous exercises on Chapter 3 · Q5
NCERT Miscellaneous Exercise Q6Miscellaneous exercises on Chapter 3 · Q6
NCERT Miscellaneous Exercise Q7Miscellaneous exercises on Chapter 3 · Q7
NCERT Miscellaneous Exercise Q8Miscellaneous exercises on Chapter 3 · Q8
NCERT Miscellaneous Exercise Q9Miscellaneous exercises on Chapter 3 · Q9
NCERT Miscellaneous Exercise Q10Miscellaneous exercises on Chapter 3 · Q10

Quick check: MCQs on Trigonometric Functions

  1. 3.1 Introduction2 questions
  2. 3.2.4 Relation between degree and radian4 questions
  3. 3.3 Trigonometric Functions3 questions
  4. 3.3.1 Sign of trigonometric functions3 questions
  5. 3.3.2 Domain and range of trigonometric functions3 questions
  6. Tangent and cotangent of a sum4 questions
  7. Double and triple angles3 questions
  8. Sums into products, products into sums3 questions
  9. Miscellaneous examples3 questions
Glossary · 13

Key terms (English–Hindi)

allied angles
Angles such as π2±x, π±x and 2π−x, whose trigonometric values are those of x up to sign, with sine and cosine swapped at π2±x.First seen in the chapter →
angleकोणkon
The amount of turning that takes a ray from its initial side to its terminal side about the vertex; anticlockwise turns are positive.First seen in the chapter →
asymptoteअनंतस्पर्शीanantasparshi
A line that a graph approaches ever more closely without meeting it, such as x=π2 for y=tan⁡x.First seen in the chapter →
degreeअंशansh
One 360th of a full turn, written 1∘; 1∘=60′ and 1′=60′′.First seen in the chapter →
even functionसम फलनsam phalan
A function with f(−x)=f(x) for every x, such as cos⁡x.First seen in the chapter →
odd functionविषम फलनvisham phalan
A function with f(−x)=−f(x) for every x, such as sin⁡x and tan⁡x.First seen in the chapter →
periodic functionआवर्ती फलनaavarti phalan
A function with f(x+p)=f(x) for all x, for some fixed p>0; the least such p is the period.First seen in the chapter →
quadrantal angle
An angle that is a whole-number multiple of π2, so that its terminal side lies along an axis.First seen in the chapter →
radianरेडियनradian
The angle at the centre of a circle cut off by an arc as long as the radius; a full turn is 2π radians.
sum-to-product formulas
Formulas that turn a sum or difference of two sines or cosines into a product, such as cos⁡x+cos⁡y=2cos⁡x+y2cos⁡x−y2.First seen in the chapter →
trigonometric functionत्रिकोणमितीय फलनtrikonmitiya phalan
A function such as sin⁡x or cos⁡x of a real number x, read from the point at arc length x on the unit circle.
trigonometric identityत्रिकोणमितीय सर्वसमिकाtrikonmitiya sarvasamika
An equation in trigonometric functions that is true for every value of the variable at which both sides are defined.First seen in the chapter →
unit circleएकक वृत्तekak vritt
The circle of radius 1 with its centre at the origin.First seen in the chapter →