Class 11Mathematics · TrigonometryFull chapter

Trigonometric Functions

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Angles and Their Measurement

Quick answer An angle is measured either in degrees (1 complete rotation = 360°) or in radians (1 radian = angle subtended by an arc equal to the radius); the two systems are linked by 180° = π radians.

An angle is formed by the rotation of a ray from an initial side to a terminal side about a fixed point called the vertex. If the rotation is anticlockwise the angle is taken as positive, and if clockwise it is taken as negative. Angles are measured in two systems.

In the sexagesimal (degree) system, one complete rotation is divided into 360 equal parts, each called one degree (1°). Further, 1° = 60 minutes (60') and 1' = 60 seconds (60").

In the circular (radian) system, one radian is the angle subtended at the centre of a circle by an arc whose length equals the radius of the circle. Since the circumference of a circle of radius r is 2πr, a complete rotation (360°) corresponds to an arc length of 2πr, and therefore equals 2π radians. This gives the fundamental relation connecting the two systems: 180° = π radians.

Using this single relation, degree measure is converted to radian measure by multiplying by π/180, and radian measure is converted to degree measure by multiplying by 180/π. For a circle of radius r, if an arc of length l subtends an angle θ (in radians) at the centre, then l = rθ; the area of the corresponding sector is A = ½ r²θ.

Example: Convert 240° into radian measure, and find the length of the arc it subtends in a circle of radius 21 cm (take π = 22/7).

  1. Radian measure = 240 × (π/180) = 240π/180 = 4π/3 radians.
  2. Here θ = 4π/3 rad and r = 21 cm.
  3. Arc length l = rθ = 21 × 4π/3 = 28π cm.
  4. Taking π = 22/7, l = 28 × 22/7 = 88 cm.

So 240° = 4π/3 radians, and the corresponding arc length is 88 cm.

Degree–Radian relation 180° = π radians the master relation for all conversions
Degree to Radian radian measure = degree measure × π/180
Radian to Degree degree measure = radian measure × 180/π
Arc length l = rθ θ must be in radians
Area of a sector A = ½ r²θ θ in radians
Degree subdivisions 1° = 60', 1' = 60"
Remember
  • π radians = 180° exactly; every degree–radian conversion uses only this one relation.
  • Radian measure is a real number (arc/radius), which is why sin x, cos x, etc. can be defined for any real x, not just for angles.
  • The formulas l = rθ and A = ½ r²θ are valid only when θ is measured in radians.
  • 1° = 60', 1' = 60"; degree subdivisions form a base-60 (sexagesimal) system.

Trigonometric Functions of Real Numbers: Quadrant Signs, Domain and Range

Quick answer Using the unit circle, sin x and cos x are defined for every real x with range [-1,1]; the ASTC rule fixes the sign of each function in every quadrant, and the remaining four functions inherit restricted domains and ranges.

Take a unit circle (radius 1) centred at the origin, and let a point P start at (1, 0) and move a directed distance x along the circle (anticlockwise if x > 0, clockwise if x < 0). If P has coordinates (a, b), define cos x = a and sin x = b. Since x can be any real number, sine and cosine are functions defined on the whole of R. The remaining four functions are defined from these two: tan x = sin x/cos x, cot x = cos x/sin x, sec x = 1/cos x, cosec x = 1/sin x. Because these are quotients, tan x and sec x are undefined wherever cos x = 0, and cot x and cosec x are undefined wherever sin x = 0.

The quadrant rule ("All Sin Tan Cos", read anticlockwise from quadrant I) tells us which functions are positive in each quadrant: in quadrant I all six functions are positive; in quadrant II only sin and cosec are positive; in quadrant III only tan and cot are positive; in quadrant IV only cos and sec are positive.

From a² + b² = 1 for a point on the unit circle, we get the fundamental identity sin²x + cos²x = 1, valid for every real x. Dividing through by cos²x and by sin²x gives the other two identities, 1 + tan²x = sec²x and 1 + cot²x = cosec²x.

Domain and range follow directly: sin x and cos x are defined for all real x and always lie between −1 and 1; tan x and sec x are undefined at odd multiples of π/2; cot x and cosec x are undefined at integer multiples of π; and since sec²x ≥ 1 and cosec²x ≥ 1, both |sec x| ≥ 1 and |cosec x| ≥ 1 always, so their range excludes the open interval (−1, 1). sin x, cos x, sec x and cosec x repeat every 2π; tan x and cot x repeat every π.

Example: If cos x = −3/5 and x lies in the third quadrant, find the values of the other five trigonometric functions.

  1. sin²x = 1 − cos²x = 1 − 9/25 = 16/25, so sin x = ±4/5.
  2. In the third quadrant sine is negative, so sin x = −4/5.
  3. tan x = sin x/cos x = (−4/5)/(−3/5) = 4/5.
  4. cosec x = 1/sin x = −5/4, sec x = 1/cos x = −5/3, cot x = 1/tan x = 5/4.

Note that tan x and cot x came out positive, exactly as the quadrant rule predicts for quadrant III.

Pythagorean identity sin²x + cos²x = 1
Identity (÷ by cos²x) 1 + tan²x = sec²x
Identity (÷ by sin²x) 1 + cot²x = cosec²x
Quotient/reciprocal relations tan x = sin x/cos x, cot x = cos x/sin x, sec x = 1/cos x, cosec x = 1/sin x
Domain of tan x, sec x R − {(2n+1)π/2 : n ∈ Z}
Domain of cot x, cosec x R − {nπ : n ∈ Z}
Range of sin x, cos x [−1, 1]
Range of tan x, cot x R tan x and cot x take every real value
Range of sec x, cosec x R − (−1, 1) i.e. |sec x| ≥ 1, |cosec x| ≥ 1
Remember
  • ASTC rule (All / Sin / Tan / Cos positive in quadrants I–IV) fixes the sign of every function once the quadrant of x is known.
  • sin²x+cos²x=1 and its two derived identities are the most-used tools for finding one function from another.
  • Domain exclusions: tan x, sec x undefined at (2n+1)π/2; cot x, cosec x undefined at nπ, n∈Z.
  • Range restrictions: sin x, cos x ∈ [−1,1]; tan x, cot x ∈ R; sec x, cosec x ∈ R − (−1,1).
  • sin x, cos x, sec x, cosec x have period 2π; tan x, cot x have period π.

Trigonometric Functions of Sum and Difference of Two Angles

Quick answer cos(x∓y) and sin(x±y) are the master compound-angle identities; every allied-angle result and every multiple-angle formula in the chapter is derived from them.

The compound (sum and difference) angle formulas express sin, cos and tan of x±y in terms of the trigonometric functions of x and y separately. They are the master identities of this chapter, since every multiple-angle, sub-multiple-angle and transformation formula is derived from them.

The two building-block results are cos(x−y) = cos x cos y + sin x sin y and cos(x+y) = cos x cos y − sin x sin y. Replacing x by (π/2 − x) in these gives the sine formulas: sin(x+y) = sin x cos y + cos x sin y and sin(x−y) = sin x cos y − cos x sin y. Dividing the sine formula by the cosine formula gives the tangent formulas: tan(x+y) = (tan x + tan y)/(1 − tan x tan y) and tan(x−y) = (tan x − tan y)/(1 + tan x tan y), valid whenever all the terms are defined.

Putting y = ±π/2, ±π etc. into these formulas produces the standard allied-angle results used to reduce any angle to an acute reference angle, such as sin(π/2 − x) = cos x, cos(π/2 − x) = sin x, sin(π − x) = sin x, cos(π − x) = −cos x, sin(π + x) = −sin x, and cos(π + x) = −cos x. Together with sin(−x) = −sin x and cos(−x) = cos x, these allow every trigonometric value to be reduced to one for an angle between 0 and π/2.

Example: Find the values of sin 75° and cos 105° using the compound angle formulas.

  1. Write 75° = 45° + 30°. Then sin 75° = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.
  2. Write 105° = 60° + 45°. Then cos 105° = cos 60° cos 45° − sin 60° sin 45° = (1/2)(√2/2) − (√3/2)(√2/2) = (√2 − √6)/4.

So sin 75° = (√6+√2)/4 ≈ 0.9659 and cos 105° = (√2−√6)/4 ≈ −0.2588, consistent with 105° being obtuse (cosine negative).

cos of difference cos(x−y) = cos x cos y + sin x sin y
cos of sum cos(x+y) = cos x cos y − sin x sin y
sin of sum sin(x+y) = sin x cos y + cos x sin y
sin of difference sin(x−y) = sin x cos y − cos x sin y
tan of sum tan(x+y) = (tan x + tan y)/(1 − tan x tan y)
tan of difference tan(x−y) = (tan x − tan y)/(1 + tan x tan y)
Allied angles sin(π/2−x)=cos x, cos(π/2−x)=sin x, sin(π−x)=sin x, cos(π−x)=−cos x, sin(π+x)=−sin x, cos(π+x)=−cos x
Remember
  • cos(x∓y) and sin(x±y) are the two families of formulas everything else in the chapter is built from.
  • Sign pattern: cos(x−y)=cosxcosy+sinxsiny and cos(x+y)=cosxcosy−sinxsiny (signs 'flip' relative to the angle sign); sine formulas keep the same sign as the angle sum/difference.
  • tan(x±y) formulas fail when 1 ∓ tan x tan y = 0, i.e. when x+y or x−y equals an odd multiple of π/2.
  • Allied-angle identities let any trigonometric ratio be rewritten using only an acute reference angle.

Multiple, Sub-multiple Angles and Transformation Formulas

Quick answer Setting y = x in the compound angle formulas gives the double- and triple-angle formulas; regrouping sums into products (and back) gives the transformation formulas used to simplify or prove trigonometric identities.

Putting y = x in the compound angle formulas gives the double-angle formulas: sin 2x = 2 sin x cos x, and cos 2x = cos²x − sin²x, which (using sin²x+cos²x=1) can also be written as cos 2x = 2cos²x − 1 = 1 − 2sin²x. Dividing sin 2x by cos 2x gives tan 2x = 2 tan x/(1 − tan²x). Writing 3x as 2x + x and expanding similarly gives the triple-angle formulas sin 3x = 3 sin x − 4 sin³x and cos 3x = 4 cos³x − 3 cos x.

A separate pair of results, called transformation formulas, convert a sum or difference of two sines/cosines into a product, and vice versa. Setting x + y = A and x − y = B in the compound angle formulas and adding/subtracting in pairs gives: sin A + sin B = 2 sin((A+B)/2) cos((A−B)/2), sin A − sin B = 2 cos((A+B)/2) sin((A−B)/2), cos A + cos B = 2 cos((A+B)/2) cos((A−B)/2), and cos A − cos B = −2 sin((A+B)/2) sin((A−B)/2). The reverse (product-to-sum) forms are 2 sin x cos y = sin(x+y) + sin(x−y), 2 cos x sin y = sin(x+y) − sin(x−y), 2 cos x cos y = cos(x+y) + cos(x−y), and 2 sin x sin y = cos(x−y) − cos(x+y).

Example: Prove that (sin 5x − sin 3x)/(cos 5x + cos 3x) = tan x.

  1. Numerator: sin 5x − sin 3x = 2 cos((5x+3x)/2) sin((5x−3x)/2) = 2 cos 4x sin x.
  2. Denominator: cos 5x + cos 3x = 2 cos((5x+3x)/2) cos((5x−3x)/2) = 2 cos 4x cos x.
  3. Ratio = (2 cos 4x sin x)/(2 cos 4x cos x) = sin x/cos x = tan x, provided cos 4x ≠ 0.

Hence the identity is proved.

Double angle (sine) sin 2x = 2 sin x cos x
Double angle (cosine) cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x
Double angle (tangent) tan 2x = 2 tan x/(1 − tan²x)
Triple angle (sine) sin 3x = 3 sin x − 4 sin³x
Triple angle (cosine) cos 3x = 4 cos³x − 3 cos x
Sum to product sinA±sinB, cosA±cosB sinA+sinB=2 sin((A+B)/2) cos((A−B)/2); sinA−sinB=2 cos((A+B)/2) sin((A−B)/2); cosA+cosB=2 cos((A+B)/2) cos((A−B)/2); cosA−cosB=−2 sin((A+B)/2) sin((A−B)/2)
Product to sum 2 sinx cosy, 2 cosx siny, 2 cosx cosy, 2 sinx siny 2sinxcosy=sin(x+y)+sin(x−y); 2cosxsiny=sin(x+y)−sin(x−y); 2cosxcosy=cos(x+y)+cos(x−y); 2sinxsiny=cos(x−y)−cos(x+y)
Remember
  • Double/triple angle formulas are just the compound angle formulas with y = x (or 3x = 2x + x); no separate derivation needs to be memorised independently.
  • cos 2x has three interchangeable forms — pick whichever matches the rest of the expression (all-cosine, all-sine, or mixed).
  • Sum-to-product formulas are the standard trick for proving that a sum of sines/cosines equals zero or simplifies to a single ratio.
  • Always keep (A+B)/2 and (A−B)/2 in the same order in both factors to avoid a sign error.

Trigonometric Equations: General Solutions

Quick answer Every basic trigonometric equation is solved by reducing it to sin x = sin a, cos x = cos a or tan x = tan a, and then applying the corresponding general-solution formula with an arbitrary integer n.

A trigonometric equation is an equation involving trigonometric functions of an unknown angle. Solutions lying in [0, 2π) are called principal solutions, while the complete set of solutions, expressed with an arbitrary integer n, is called the general solution.

Three master results generate the general solution of every basic trigonometric equation: if sin x = sin a then x = nπ + (−1)ⁿ a; if cos x = cos a then x = 2nπ ± a; and if tan x = tan a then x = nπ + a, in every case n ∈ Z and a is the principal value satisfying the equation. Two special cases follow at once: sin x = 0 gives x = nπ, and cos x = 0 gives x = (2n+1)π/2. Also, if sin²x = sin²a (or the analogous statement for cos² or tan²), the general solution is simply x = nπ ± a.

The general method is: (i) use identities to reduce the equation to one of these standard forms (often after factorising or solving a quadratic in sin x, cos x or tan x), (ii) find the principal value a, and (iii) write the corresponding general solution, discarding any values outside the natural range of the function (for example, sin x or cos x can never exceed 1 in magnitude).

Example: Solve tan²x = 3.

  1. tan²x = 3 = (√3)² = tan²(π/3), since tan(π/3) = √3.
  2. Using the result for tan²x = tan²a, the general solution is x = nπ ± π/3, n ∈ Z.
  3. Check: at x = π/3, tan x = √3 so tan²x = 3 ✓; at x = −π/3, tan x = −√3 so tan²x = 3 ✓.

So the complete solution set is x = nπ ± π/3, n ∈ Z.

sin x = sin a x = nπ + (−1)ⁿ a, n ∈ Z
cos x = cos a x = 2nπ ± a, n ∈ Z
tan x = tan a x = nπ + a, n ∈ Z
sin x = 0 x = nπ, n ∈ Z
cos x = 0 x = (2n+1)π/2, n ∈ Z
sin²x = sin²a (also cos², tan²) x = nπ ± a, n ∈ Z
Remember
  • Always reduce the given equation to sin x = sin a, cos x = cos a or tan x = tan a form before applying a general-solution formula.
  • The integer n ranges over all of Z; different values of n generate every solution, not just those in one interval.
  • Reject any algebraic solution outside the natural range of the function, e.g. sin x = 2 has no solution.
  • For sin²x = sin²a, cos²x = cos²a or tan²x = tan²a, the compact general solution is always x = nπ ± a.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

180° = π radians
Degree–Radian relation
radian measure = degree measure × π/180
Degree to Radian
degree measure = radian measure × 180/π
Radian to Degree
l = rθ
Arc length
A = ½ r²θ
Area of a sector
1° = 60', 1' = 60"
Degree subdivisions
sin²x + cos²x = 1
Pythagorean identity
1 + tan²x = sec²x
Identity (÷ by cos²x)
1 + cot²x = cosec²x
Identity (÷ by sin²x)
tan x = sin x/cos x, cot x = cos x/sin x, sec x = 1/cos x, cosec x = 1/sin x
Quotient/reciprocal relations
R − {(2n+1)π/2 : n ∈ Z}
Domain of tan x, sec x
R − {nπ : n ∈ Z}
Domain of cot x, cosec x
[−1, 1]
Range of sin x, cos x
R
Range of tan x, cot x
R − (−1, 1)
Range of sec x, cosec x
cos(x−y) = cos x cos y + sin x sin y
cos of difference
cos(x+y) = cos x cos y − sin x sin y
cos of sum
sin(x+y) = sin x cos y + cos x sin y
sin of sum
sin(x−y) = sin x cos y − cos x sin y
sin of difference
tan(x+y) = (tan x + tan y)/(1 − tan x tan y)
tan of sum
tan(x−y) = (tan x − tan y)/(1 + tan x tan y)
tan of difference
sin(π/2−x)=cos x, cos(π/2−x)=sin x, sin(π−x)=sin x, cos(π−x)=−cos x, sin(π+x)=−sin x, cos(π+x)=−cos x
Allied angles
sin 2x = 2 sin x cos x
Double angle (sine)
cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x
Double angle (cosine)
tan 2x = 2 tan x/(1 − tan²x)
Double angle (tangent)
sin 3x = 3 sin x − 4 sin³x
Triple angle (sine)
cos 3x = 4 cos³x − 3 cos x
Triple angle (cosine)
sinA±sinB, cosA±cosB
Sum to product
2 sinx cosy, 2 cosx siny, 2 cosx cosy, 2 sinx siny
Product to sum
x = nπ + (−1)ⁿ a, n ∈ Z
sin x = sin a
x = 2nπ ± a, n ∈ Z
cos x = cos a
x = nπ + a, n ∈ Z
tan x = tan a
x = nπ, n ∈ Z
sin x = 0
x = (2n+1)π/2, n ∈ Z
cos x = 0
x = nπ ± a, n ∈ Z
sin²x = sin²a (also cos², tan²)

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Angle measurement easy

5π/6 radians is equal to which degree measure?

Q2 Angle measurement easy

The radian measure corresponding to 25° is

Q3 Arc length medium

A circle has radius 10 cm. Find the length of the arc that subtends an angle of 30° at the centre.

Q4 Quadrant signs easy

In which quadrant does 210° lie, and what is the sign of sin 210°?

Q5 Domain and range easy

The domain of tan x is

Q6 Domain and range medium

The range of sec x is

Q7 Sum and difference formulas medium

Using the compound angle formula, sin 75° equals

Q8 Sum and difference formulas hard

The value of tan 15° is

Q9 Multiple angle formulas medium

If sin x = 3/5 and 0 < x < π/2, then sin 2x equals

Q10 Trigonometric equations medium

The general solution of cos x = 1/2 is

Q11 Trigonometric equations medium

The general solution of tan x = −1 is

Q12 Transformation formulas medium

sin 4x − sin 2x is equal to

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Find the radian measure corresponding to the degree measure 520°.Angle measurement

Use the relation 180° = π radians, so degree measure is converted to radian measure by multiplying by π/180.

  1. Radian measure = 520 × π/180.
  2. 520/180 = 26/9 (dividing numerator and denominator by 20).
  3. So radian measure = 26π/9.

Hence 520° = 26π/9 radians.

2 Find the degree measure corresponding to the radian measure 11π/16.Angle measurement

Use radian measure × 180/π to convert to degree measure.

  1. Degree measure = (11π/16) × (180/π) = 11 × 180/16 = 1980/16.
  2. 1980/16 = 123.75°.
  3. The fractional part 0.75° = 0.75 × 60' = 45'.

Hence 11π/16 radians = 123°45'.

3 If cot x = −5/12 and x lies in the second quadrant, find the values of sin x, cos x, tan x, sec x and cosec x.Signs, domain and range

In the second quadrant, sine and cosecant are positive while cosine, tangent and secant are negative.

  1. tan x = 1/cot x = 1/(−5/12) = −12/5.
  2. Using the 5–12–13 Pythagorean triple as a reference, the magnitudes are sin x = 12/13, cos x = 5/13, tan x = 12/5.
  3. Applying the quadrant-II signs: sin x = 12/13 (positive), cos x = −5/13 (negative), tan x = −12/5 (negative, matching step 1).
  4. cosec x = 1/sin x = 13/12, sec x = 1/cos x = −13/5.

So sin x = 12/13, cos x = −5/13, tan x = −12/5, cosec x = 13/12, sec x = −13/5.

4 Prove that cos(π/4 − x) cos(π/4 − y) − sin(π/4 − x) sin(π/4 − y) = sin(x + y).Sum and difference formulas

The left side has the exact form cos A cos B − sin A sin B, which equals cos(A + B), with A = π/4 − x and B = π/4 − y.

  1. A + B = (π/4 − x) + (π/4 − y) = π/2 − (x + y).
  2. So the left side equals cos(π/2 − (x + y)).
  3. Using the allied-angle identity cos(π/2 − θ) = sin θ with θ = x + y, this equals sin(x + y).

Hence the left side equals the right side, and the identity is proved.

5 Find the general solution of the equation 2cos²x + 3 sin x = 0.Trigonometric equations

Replace cos²x by 1 − sin²x so that the equation is entirely in terms of sin x.

  1. 2(1 − sin²x) + 3 sin x = 0 ⟹ 2 − 2sin²x + 3 sin x = 0 ⟹ 2sin²x − 3 sin x − 2 = 0.
  2. Treat this as a quadratic in sin x: sin x = [3 ± √(9 + 16)]/4 = [3 ± 5]/4, giving sin x = 2 or sin x = −1/2.
  3. Since sin x = 2 is impossible (sin x ∈ [−1, 1]), only sin x = −1/2 = sin(−π/6) is valid.
  4. Using sin x = sin a with a = −π/6, the general solution is x = nπ + (−1)ⁿ(−π/6), n ∈ Z, which is the same as x = nπ + (−1)ⁿ⁺¹ π/6.

Hence the general solution is x = nπ + (−1)ⁿ⁺¹ (π/6), n ∈ Z.

6 Prove that (cos x − cos y)² + (sin x − sin y)² = 4 sin²((x−y)/2).Multiple angle formulas

Expand both squares and use the Pythagorean identity.

  1. (cos x − cos y)² + (sin x − sin y)² = cos²x − 2cos x cos y + cos²y + sin²x − 2 sin x sin y + sin²y.
  2. Group as (sin²x + cos²x) + (sin²y + cos²y) − 2(cos x cos y + sin x sin y) = 1 + 1 − 2 cos(x − y) = 2 − 2 cos(x − y).
  3. Using cos(x−y) = 1 − 2 sin²((x−y)/2), we get 2 − 2[1 − 2sin²((x−y)/2)] = 4 sin²((x−y)/2).

Hence the identity is proved.

Previous-year board questions 4

Q1 Find the radian measure of 40°20'. CBSE 2020 1 mark

First express 40°20' as a pure degree value.

  1. 40°20' = 40° + 20/60° = 40° + 1/3° = 121/3 degrees.
  2. Radian measure = (121/3) × (π/180) = 121π/540.

Hence 40°20' = 121π/540 radians.

Q2 Prove that tan 3x tan 2x tan x = tan 3x − tan 2x − tan x. CBSE 2019 3 marks

Write 3x = 2x + x and use the tangent sum formula.

  1. tan 3x = tan(2x + x) = (tan 2x + tan x)/(1 − tan 2x tan x).
  2. Cross-multiplying: tan 3x (1 − tan 2x tan x) = tan 2x + tan x.
  3. Expanding: tan 3x − tan 3x tan 2x tan x = tan 2x + tan x.
  4. Rearranging: tan 3x − tan 2x − tan x = tan 3x tan 2x tan x.

This is exactly the required identity, so tan 3x tan 2x tan x = tan 3x − tan 2x − tan x.

Q3 Prove that sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x. CBSE 2018 3 marks

Pair the outer and inner terms and apply the sum-to-product formula to each pair.

  1. (sin x + sin 7x) = 2 sin 4x cos 3x, since (x+7x)/2 = 4x and (7x−x)/2 = 3x.
  2. (sin 3x + sin 5x) = 2 sin 4x cos x, since (3x+5x)/2 = 4x and (5x−3x)/2 = x.
  3. Adding: sin x + sin 3x + sin 5x + sin 7x = 2 sin 4x cos 3x + 2 sin 4x cos x = 2 sin 4x (cos 3x + cos x).
  4. Now cos 3x + cos x = 2 cos 2x cos x, so the expression becomes 2 sin 4x × 2 cos 2x cos x = 4 cos x cos 2x sin 4x.

Hence the identity is proved.

Q4 Solve the equation sin x + sin 3x + sin 5x = 0, and find its general solution. CBSE 2023 5 marks

Group the first and third terms, which are symmetric about the middle term.

  1. sin x + sin 5x = 2 sin 3x cos 2x, since (x+5x)/2 = 3x and (5x−x)/2 = 2x.
  2. So the equation becomes 2 sin 3x cos 2x + sin 3x = 0, i.e. sin 3x (2 cos 2x + 1) = 0.
  3. Case 1: sin 3x = 0 ⟹ 3x = nπ ⟹ x = nπ/3, n ∈ Z.
  4. Case 2: cos 2x = −1/2 = cos(2π/3) ⟹ 2x = 2nπ ± 2π/3 ⟹ x = nπ ± π/3, n ∈ Z.

Hence the general solution is x = nπ/3, n ∈ Z, together with x = nπ ± π/3, n ∈ Z.

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