Solved Examples

How to Use This Section

Here are 32 fully worked problems covering the entire chapter — angle conversions and arc lengths, unit-circle evaluations, sign and quadrant work, sum-difference identities, multiple angles and the product-sum toolkit, including the heavyweight miscellaneous-level identities — arranged in a deliberate easy → medium → hard progression.

One suggestion that multiplies the value of every problem: attempt each yourself before reading the solution. For identity proofs, write your own line of attack first (which side to start from, which formula family to use) and compare strategies, not just answers. Problems 1-10 warm up on measures and values, 11-22 build identity fluency, 23-32 are Board and JEE-level proofs.

Solved Examples

Example 1: Conversion warm-up

Convert (i) 75° to radians (ii) $\frac{7\pi}{6}$ to degrees (iii) 2 radians to degrees (approximately).

Solution:

Step 1 — (i) Degrees to radians: multiply by $\frac{\pi}{180}$. $75° = 75 \times \frac{\pi}{180} = \frac{5\pi}{12}$ radian.

Step 2 — (ii) Radians to degrees: multiply by $\frac{180°}{\pi}$. $\frac{7\pi}{6} = \frac{7 \times 180°}{6} = 210°$.

Step 3 — (iii) Decimal estimate. 1 radian $\approx 57.3°$, so 2 radians $\approx 114.6°$.

Takeaway: Exact fractions of $\pi$ for (i)-(ii); the 57.3° estimate for quick decimal work.

Example 2: Arc length both ways

A circular wire of radius 3 cm is cut and bent along a circle of radius 48 cm. Find the angle it subtends at the centre.

Solution:

Step 1 — Find the length of the wire. The wire is a full circle of radius 3, so its length is the circumference $2\pi \times 3 = 6\pi$ cm.

Step 2 — Treat that length as an arc of the big circle. $\theta = \frac{l}{r} = \frac{6\pi}{48} = \frac{\pi}{8}$ radian.

Step 3 — Convert if degrees are wanted. $\frac{\pi}{8} = \frac{180°}{8} = 22.5°$.

Takeaway: A classic two-step: compute the arc length from one circle, divide by the radius of the other.

Example 3: Five functions from tan

If $\tan x = -\frac{5}{12}$ and x lies in the second quadrant, find sin x and cos x.

Solution:

Step 1 — Go through the companion identity. $\sec^2 x = 1 + \tan^2 x = 1 + \frac{25}{144} = \frac{169}{144}$, so $\sec x = \pm\frac{13}{12}$.

Step 2 — Sign from the quadrant. In Q II cosine (and sec) are negative: $\sec x = -\frac{13}{12}$, hence $\cos x = -\frac{12}{13}$.

Step 3 — Recover sine as a product. $\sin x = \tan x \cdot \cos x = \left(-\frac{5}{12}\right)\left(-\frac{12}{13}\right) = \frac{5}{13}$ — positive, exactly as Q II demands. ✓

Takeaway: The 5-12-13 triangle plus quadrant signs — the standard machine for completing a function set.

Example 4: Big-angle evaluation set

Evaluate (i) $\sin\frac{17\pi}{6}$ (ii) $\tan\left(-\frac{7\pi}{4}\right)$ (iii) $\sec 420°$.

Solution:

Step 1 — (i) Strip a revolution. $\frac{17\pi}{6} = 2\pi + \frac{5\pi}{6}$, so $\sin\frac{17\pi}{6} = \sin\frac{5\pi}{6} = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\frac{\pi}{6} = \frac{1}{2}$.

Step 2 — (ii) Odd rule, then reduce. $\tan\left(-\frac{7\pi}{4}\right) = -\tan\frac{7\pi}{4} = -\tan\left(2\pi - \frac{\pi}{4}\right) = -\left(-\tan\frac{\pi}{4}\right) = 1$.

Step 3 — (iii) Strip 360°. $420° = 360° + 60°$, so $\sec 420° = \sec 60° = 2$.

Takeaway: Strip revolutions, use allied angles, track signs — three moves cover every evaluation.

Example 5: Identity from the fundamental trio

Prove that $\sec^2 x + \csc^2 x = \sec^2 x \csc^2 x$.

Solution:

Step 1 — Drop to sines and cosines. LHS $= \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}$.

Step 2 — Combine over a common denominator. $= \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x}$.

Step 3 — Apply the fundamental identity. The numerator is 1: $= \frac{1}{\sin^2 x \cos^2 x} = \sec^2 x\csc^2 x$. ∎

Takeaway: Whenever secs and cosecs mix, drop to sines and cosines — the fundamental identity is usually waiting one line down.

Example 6: Quadrant from double conditions

If $\tan x > 0$ and $\sec x < 0$, find the quadrant of x, and the sign of $\sin\frac{x}{2}$ if additionally $\pi < x < \frac{3\pi}{2}$.

Solution:

Step 1 — List each condition's quadrants from the ASTC chart. tan positive → Q I or Q III; sec (same sign as cos) negative → Q II or Q III.

Step 2 — Intersect. The only common quadrant is Q III.

Step 3 — Halve the interval for the half angle. $\pi < x < \frac{3\pi}{2}$ gives $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$ — Q II.

Step 4 — Read the sign. In Q II sine is positive: $\sin\frac{x}{2} > 0$.

Takeaway: Halving an angle halves its interval — re-locate the quadrant of $\frac{x}{2}$ explicitly; never assume it matches x.

Example 7: The 2-sin-cos completing trick

If $\sin x + \cos x = \frac{1}{2}$, find $\sin x \cos x$ and $\sin^3 x + \cos^3 x$.

Solution:

Step 1 — Square the given equation. $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x\cos x = 1 + 2\sin x\cos x = \frac{1}{4}$.

Step 2 — Solve for the product. $2\sin x\cos x = \frac{1}{4} - 1 = -\frac{3}{4}$, so $\sin x\cos x = -\frac{3}{8}$.

Step 3 — Use the cube identity. $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ with $a + b = \frac{1}{2}$, $ab = -\frac{3}{8}$.

Step 4 — Substitute. $\left(\frac{1}{2}\right)^3 - 3\left(-\frac{3}{8}\right)\frac{1}{2} = \frac{1}{8} + \frac{9}{16} = \frac{11}{16}$.

Takeaway: $(\sin x + \cos x)^2 = 1 + \sin 2x$ — squaring the sum unlocks every symmetric combination.

Example 8: Allied-angle evaluation chain

Evaluate $\sin 780° \sin 480° + \cos 120° \cos 60°$.

Solution:

Step 1 — Reduce each sine. $\sin 780° = \sin(720° + 60°) = \sin 60° = \frac{\sqrt{3}}{2}$; $\sin 480° = \sin(360° + 120°) = \sin 120° = \sin(180° - 60°) = \frac{\sqrt{3}}{2}$.

Step 2 — Write the cosines. $\cos 120° = \cos(180° - 60°) = -\frac{1}{2}$; $\cos 60° = \frac{1}{2}$.

Step 3 — Assemble. $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \left(-\frac{1}{2}\right)\frac{1}{2} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$.

Takeaway: Reduce each factor independently, then combine — never mix reduction and arithmetic in one step.

Example 9: A √2 extraction

Prove that $\sin x + \cos x = \sqrt{2}\cos\left(x - \frac{\pi}{4}\right)$.

Solution:

Step 1 — Expand the right side. $\sqrt{2}\cos\left(x - \frac{\pi}{4}\right) = \sqrt{2}\left[\cos x\cos\frac{\pi}{4} + \sin x \sin\frac{\pi}{4}\right]$.

Step 2 — Substitute the values. $= \sqrt{2} \cdot \frac{1}{\sqrt{2}}(\cos x + \sin x) = \sin x + \cos x$ = LHS. ∎

Takeaway: Factoring $\sqrt{2}$ out of $\sin x + \cos x$ is the baby case of the $a\sin x + b\cos x = R\sin(x + \phi)$ technique in the JEE Corner.

Example 10: Value of cos(x + y)cos(x − y)

Prove that $\cos(x + y)\cos(x - y) = \cos^2 x - \sin^2 y$.

Solution:

Step 1 — Expand as a difference of squares. $(\cos x\cos y - \sin x\sin y)(\cos x\cos y + \sin x\sin y) = \cos^2 x\cos^2 y - \sin^2 x\sin^2 y$.

Step 2 — Convert everything toward x-cosines and y-sines. $= \cos^2 x(1 - \sin^2 y) - (1 - \cos^2 x)\sin^2 y$.

Step 3 — Expand and cancel. $= \cos^2 x - \cos^2 x\sin^2 y - \sin^2 y + \cos^2 x\sin^2 y = \cos^2 x - \sin^2 y$. ∎

Takeaway: The difference-of-squares shape $(A-B)(A+B)$ collapses the product before any heavy algebra.

Example 11: The 2cos(x/2) family flagship

Prove that $\cos x + \cos y + \cos z + \cos(x + y + z) = 4\cos\frac{x+y}{2}\cos\frac{y+z}{2}\cos\frac{z+x}{2}$.

Solution:

Step 1 — Pair strategically. Group as $[\cos x + \cos y] + [\cos z + \cos(x+y+z)]$ — chosen so both pairs will share a factor.

Step 2 — Sum to product on each pair. $2\cos\frac{x+y}{2}\cos\frac{x-y}{2} + 2\cos\frac{x+y+2z}{2}\cos\frac{x+y}{2}$.

Step 3 — Factor the common half-sum. $= 2\cos\frac{x+y}{2}\left[\cos\frac{x-y}{2} + \cos\frac{x+y+2z}{2}\right]$.

Step 4 — Sum to product again inside the bracket. Half-sum: $\frac{1}{2}\left[\frac{x-y}{2} + \frac{x+y+2z}{2}\right] = \frac{x+z}{2}$; half-difference: $\frac{y+z}{2}$. The bracket becomes $2\cos\frac{x+z}{2}\cos\frac{y+z}{2}$.

Step 5 — Multiply out. Total $= 4\cos\frac{x+y}{2}\cos\frac{y+z}{2}\cos\frac{z+x}{2}$. ∎

Takeaway: Pair the terms so both pairs share a common half-sum factor; two rounds of sum-to-product finish it — a Board favourite.

Example 12: Sum of two squared pair-sums

Prove that $(\cos x + \cos y)^2 + (\sin x + \sin y)^2 = 4\cos^2\frac{x - y}{2}$.

Solution:

Step 1 — Expand both squares and add. The four squared terms give $(\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) = 2$; the cross terms give $2(\cos x\cos y + \sin x\sin y)$.

Step 2 — Recognise the difference formula. $\cos x\cos y + \sin x\sin y = \cos(x - y)$, so the total is $2 + 2\cos(x - y)$.

Step 3 — Half-angle square. $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$ with $\theta = x - y$: $2[1 + \cos(x-y)] = 4\cos^2\frac{x-y}{2}$. ∎

Takeaway: Sums of squares of paired sines/cosines almost always reduce to $2 + 2\cos(\text{gap})$ — then $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$ lands the finish.

Example 13: A three-angle telescoping product

Prove the classic: $\sin 20° \sin 40° \sin 60° \sin 80° = \frac{3}{16}$.

Solution:

Step 1 — Peel off the known value. $\sin 60° = \frac{\sqrt{3}}{2}$, so the product is $\frac{\sqrt{3}}{2}\left[\sin 20°\sin 40°\sin 80°\right]$.

Step 2 — Recognise the triple-angle product family. $\sin\theta \sin(60° - \theta)\sin(60° + \theta) = \frac{1}{4}\sin 3\theta$; with $\theta = 20°$ the three factors are exactly sin 20°, sin 40°, sin 80°.

Step 3 — Apply it. $\sin 20°\sin 40°\sin 80° = \frac{1}{4}\sin 60° = \frac{\sqrt{3}}{8}$.

Step 4 — Multiply. $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{8} = \frac{3}{16}$. ∎

Takeaway: The triple-angle product identity $\sin\theta\sin(60°-\theta)\sin(60°+\theta) = \frac{\sin 3\theta}{4}$ (and its cos twin) crack all products over 20°-40°-80°-type families.

Example 14: tan sum with a condition

If $x + y = \frac{\pi}{4}$, prove that $(1 + \tan x)(1 + \tan y) = 2$.

Solution:

Step 1 — Take tan of the condition. $\tan(x + y) = \tan\frac{\pi}{4} = 1$, so $\frac{\tan x + \tan y}{1 - \tan x\tan y} = 1$.

Step 2 — Cross-multiply. $\tan x + \tan y = 1 - \tan x\tan y$.

Step 3 — Expand the target product. $(1 + \tan x)(1 + \tan y) = 1 + \tan x + \tan y + \tan x\tan y$.

Step 4 — Substitute Step 2. $= 1 + (1 - \tan x\tan y) + \tan x\tan y = 2$. ∎

Takeaway: Conditions on x + y convert to relations between tan x and tan y through the sum formula — then it is pure algebra.

Example 15: Half-angle values from cos x

If $\cos x = -\frac{1}{3}$ with $\pi < x < \frac{3\pi}{2}$, find $\sin\frac{x}{2}$, $\cos\frac{x}{2}$ and $\tan\frac{x}{2}$.

Solution:

Step 1 — Locate the half angle FIRST. $\pi < x < \frac{3\pi}{2}$ gives $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$ — Q II: sin positive, cos negative, tan negative.

Step 2 — Power-reduce for sine. $\sin^2\frac{x}{2} = \frac{1 - \cos x}{2} = \frac{1 + \frac{1}{3}}{2} = \frac{2}{3}$, so $\sin\frac{x}{2} = +\sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$.

Step 3 — Power-reduce for cosine. $\cos^2\frac{x}{2} = \frac{1 + \cos x}{2} = \frac{1 - \frac{1}{3}}{2} = \frac{1}{3}$, and Q II makes it negative: $\cos\frac{x}{2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$.

Step 4 — Divide for tangent. $\tan\frac{x}{2} = \frac{\sqrt{6}/3}{-\sqrt{3}/3} = -\sqrt{2}$ — negative in Q II. ✓

Takeaway: Locate the half-angle's quadrant FIRST, then power-reduce and attach signs.

Example 16: The cot x cot 2x identity

Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$.

Solution:

Step 1 — Write 3x as a sum with cot. $\cot 3x = \cot(2x + x) = \frac{\cot 2x\cot x - 1}{\cot 2x + \cot x}$.

Step 2 — Cross-multiply. $\cot 3x(\cot 2x + \cot x) = \cot 2x\cot x - 1$, i.e. $\cot 3x\cot 2x + \cot 3x\cot x = \cot 2x\cot x - 1$.

Step 3 — Rearrange to the target. $\cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x = 1$. ∎

Takeaway: The cot-sum formula cross-multiplied — mirror twin of the tan 3A identity.

Example 17: Four-term sum-to-product

Prove that $\frac{\sin x + \sin 3x + \sin 5x + \sin 7x}{\cos x + \cos 3x + \cos 5x + \cos 7x} = \tan 4x$.

Solution:

Step 1 — Pair outers with outers, inners with inners. $[\sin 7x + \sin x] + [\sin 5x + \sin 3x]$ — both pairs have half-sum 4x.

Step 2 — Factor the numerator. $2\sin 4x\cos 3x + 2\sin 4x\cos x = 2\sin 4x(\cos 3x + \cos x)$.

Step 3 — Same pairing below. $\cos x + \cos 3x + \cos 5x + \cos 7x = 2\cos 4x\cos 3x + 2\cos 4x\cos x = 2\cos 4x(\cos 3x + \cos x)$.

Step 4 — Cancel the common bracket. Quotient $= \frac{\sin 4x}{\cos 4x} = \tan 4x$. ∎

Takeaway: With four terms, pair so both pairs produce the SAME half-sum (here 4x) — outer-with-outer, inner-with-inner.

Example 18: A conditional identity with x + y

If $\sin x + \sin y = a$ and $\cos x + \cos y = b$, find $\tan\frac{x+y}{2}$.

Solution:

Step 1 — Convert both data to products. $a = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$ and $b = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$.

Step 2 — Divide to kill the common factor. $\frac{a}{b} = \frac{\sin\frac{x+y}{2}}{\cos\frac{x+y}{2}} = \tan\frac{x+y}{2}$ (for $b \neq 0$).

Takeaway: Dividing the two sum-to-product forms kills the common half-difference factor — the half-sum's tangent pops out free.

Example 19: Range preview with identities

Find the maximum value of $\cos^2 x - \sin^2 x + 1$ over all real x.

Solution:

Step 1 — Recognise the double angle. $\cos^2 x - \sin^2 x = \cos 2x$, so the expression is $\cos 2x + 1$.

Step 2 — Bound it. $\cos 2x \in [-1, 1]$ gives $\cos 2x + 1 \in [0, 2]$.

Step 3 — Attainment. The maximum 2 occurs when $\cos 2x = 1$, i.e. at $x = n\pi$.

Takeaway: Identities convert ugly quadratic-in-sin/cos expressions into single trig functions whose range is known on sight.

Example 20: Half-angle set from tan x

If $\tan x = -\frac{4}{3}$ with x in quadrant II, find $\sin\frac{x}{2}$, $\cos\frac{x}{2}$ and $\tan\frac{x}{2}$.

Solution:

Step 1 — Recover cos x with the quadrant sign. $\sec^2 x = 1 + \frac{16}{9} = \frac{25}{9}$; Q II makes sec negative: $\sec x = -\frac{5}{3}$, so $\cos x = -\frac{3}{5}$.

Step 2 — Locate the half angle. $\frac{\pi}{2} < x < \pi$ gives $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ — Q I, where ALL functions are positive.

Step 3 — Power-reduce for sine. $\sin\frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$ — note $1 - \cos x = 1 + \frac{3}{5}$ because cos x is negative.

Step 4 — Power-reduce for cosine. $\cos\frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \frac{1}{\sqrt{5}}$.

Step 5 — Divide. $\tan\frac{x}{2} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2$.

Takeaway: Note $1 - \cos x = 1 + \frac{3}{5}$ because cos x is negative; sign slips here cost the whole question.

Example 21: An equation solved by factoring (preview)

Find all x in $[0, 2\pi]$ with $\sin 2x = \sin x$.

Solution:

Step 1 — Bring everything to one side and factor. $2\sin x\cos x - \sin x = \sin x(2\cos x - 1) = 0$.

Step 2 — Case sin x = 0. In $[0, 2\pi]$: $x = 0, \pi, 2\pi$.

Step 3 — Case $\cos x = \frac{1}{2}$. In $[0, 2\pi]$: $x = \frac{\pi}{3}, \frac{5\pi}{3}$.

Step 4 — Collect. Solution set: $\left\{0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi\right\}$.

Takeaway: NEVER divide by sin x — factoring keeps the sin x = 0 family alive. (General solutions of such equations: JEE Corner.)

Example 22: Proving via difference of squares

Prove that $\cos^2 2x - \cos^2 6x = \sin 4x \sin 8x$.

Solution:

Step 1 — Factor as a difference of squares. $(\cos 2x - \cos 6x)(\cos 2x + \cos 6x)$.

Step 2 — Sum-to-product on each factor. $\cos 2x - \cos 6x = -2\sin 4x\sin(-2x) = 2\sin 4x \sin 2x$; and $\cos 2x + \cos 6x = 2\cos 4x\cos 2x$.

Step 3 — Regroup into double angles. $\left[2\sin 4x\sin 2x\right]\left[2\cos 4x\cos 2x\right] = (2\sin 4x\cos 4x)(2\sin 2x\cos 2x) = \sin 8x \sin 4x$. ∎

Takeaway: Difference of squared cosines → factor → sum-to-product both factors → double angles reassemble.

Example 23: The cos 6x expansion

Express cos 6x in terms of cos x — outline the chain and give the result.

Solution:

Step 1 — Choose the nesting. $6x = 2(3x)$: apply the double angle outside, the triple angle inside. $\cos 6x = 2\cos^2 3x - 1$ with $\cos 3x = 4\cos^3 x - 3\cos x$.

Step 2 — Substitute. $\cos 6x = 2(4\cos^3 x - 3\cos x)^2 - 1$.

Step 3 — Expand the square. $(4\cos^3 x - 3\cos x)^2 = 16\cos^6 x - 24\cos^4 x + 9\cos^2 x$.

Step 4 — Finish. $\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$.

Step 5 — Check at x = 0. $32 - 48 + 18 - 1 = 1 = \cos 0$. ✓

Takeaway: Nest 6x = 2(3x): double-angle outside, triple-angle inside — and check the coefficients 32, −48, 18, −1 at x = 0.

Example 24: sin x sin 2x sin 3x expansion

Prove that $\sin x \sin 2x \sin 3x$ equals $\frac{1}{4}(\sin 2x + \sin 4x - \sin 6x)$.

Solution:

Step 1 — Product-to-sum on the outer pair. $\sin x\sin 3x = \frac{1}{2}[\cos 2x - \cos 4x]$.

Step 2 — Multiply by sin 2x. $\frac{1}{2}[\sin 2x\cos 2x - \sin 2x\cos 4x]$.

Step 3 — Product-to-sum on each piece. $\sin 2x\cos 2x = \frac{1}{2}\sin 4x$; and $\sin 2x\cos 4x = \frac{1}{2}[\sin 6x + \sin(-2x)] = \frac{1}{2}[\sin 6x - \sin 2x]$.

Step 4 — Assemble. $\frac{1}{4}\sin 4x - \frac{1}{4}\sin 6x + \frac{1}{4}\sin 2x = \frac{1}{4}(\sin 2x + \sin 4x - \sin 6x)$. ∎

Takeaway: Triple products fall to two rounds of product-to-sum — work outside-in, keeping every $\frac{1}{2}$.

Example 25: A cot identity with three terms

Prove that $\cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x)$.

Solution:

Step 1 — Simplify the LHS. $\frac{\cos 4x}{\sin 4x} \cdot (\sin 5x + \sin 3x) = \frac{\cos 4x}{\sin 4x} \cdot 2\sin 4x\cos x = 2\cos 4x\cos x$.

Step 2 — Simplify the RHS. $\frac{\cos x}{\sin x} \cdot (\sin 5x - \sin 3x) = \frac{\cos x}{\sin x} \cdot 2\cos 4x\sin x = 2\cos 4x\cos x$.

Step 3 — Compare. Both sides equal $2\cos 4x\cos x$. ∎

Takeaway: When both sides look messy, simplify EACH side independently to a common target — often faster than transforming one into the other.

Example 26: The sin²-difference identity

Prove that $\sin^2 6x - \sin^2 4x = \sin 2x \sin 10x$.

Solution:

Step 1 — Quote the compact identity. $\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B)$; with A = 6x, B = 4x this is $\sin 10x \sin 2x$ immediately. ∎

Step 2 — Know why it works. $\sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B)$; sum-to-product on both factors gives $2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \cdot 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$, and pairing the halves reassembles $\sin(A+B)\sin(A-B)$.

Takeaway: Memorise $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$ and its twin $\cos^2 A - \cos^2 B = -\sin(A+B)\sin(A-B)$ — a JEE one-liner.

Example 27: Conditional identity, A + B + C = π preview

If $x + y + z = \pi$, prove that $\sin 2x + \sin 2y + \sin 2z = 4\sin x\sin y\sin z$.

Solution:

Step 1 — Pair the first two. $\sin 2x + \sin 2y = 2\sin(x + y)\cos(x - y) = 2\sin z\cos(x - y)$, using $\sin(x + y) = \sin(\pi - z) = \sin z$.

Step 2 — Open the third. $\sin 2z = 2\sin z\cos z$.

Step 3 — Factor $2\sin z$. Total $= 2\sin z[\cos(x - y) + \cos z] = 2\sin z[\cos(x - y) - \cos(x + y)]$, since $\cos z = \cos(\pi - (x+y)) = -\cos(x+y)$.

Step 4 — Product form. $\cos(x-y) - \cos(x+y) = 2\sin x\sin y$, so the total is $4\sin x\sin y\sin z$. ∎

Takeaway: Under $x + y + z = \pi$: $\sin(x+y) = \sin z$ and $\cos(x+y) = -\cos z$ — substitute these early and often. Full conditional-identity kit: JEE Corner.

Example 28: Evaluating a mixed surd expression

Show that $\frac{\sin 75° - \sin 15°}{\cos 75° + \cos 15°} = \tan 30°$, and evaluate it.

Solution:

Step 1 — Factor the numerator. $\sin 75° - \sin 15° = 2\cos\frac{75° + 15°}{2}\sin\frac{75° - 15°}{2} = 2\cos 45°\sin 30°$.

Step 2 — Factor the denominator. $\cos 75° + \cos 15° = 2\cos 45°\cos 30°$.

Step 3 — Cancel and evaluate. $\frac{\sin 30°}{\cos 30°} = \tan 30° = \frac{1}{\sqrt{3}}$. ∎

Takeaway: Even with concrete angles, sum-to-product beats plugging in surd values — cleaner and error-proof.

Example 29: The mixed-sign square sum

Prove that $(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2\frac{x+y}{2}$.

Solution:

Step 1 — Expand and add. Squared terms total 2; cross terms give $2(\cos x\cos y - \sin x\sin y)$.

Step 2 — Recognise the SUM formula. $\cos x\cos y - \sin x\sin y = \cos(x + y)$, so the total is $2 + 2\cos(x + y)$.

Step 3 — Half-angle square. $2[1 + \cos(x+y)] = 4\cos^2\frac{x+y}{2}$. ∎

Takeaway: Compare with Example 12: a PLUS pairing gives the half-DIFFERENCE, a mixed pairing gives the half-SUM — the sign pattern in the squares decides which.

Example 30: An 8-fold angle chain

Prove that $\cos x \cos 2x \cos 4x \cos 8x = \frac{\sin 16x}{16\sin x}$ (for $\sin x \neq 0$).

Solution:

Step 1 — Multiply and divide by $2\sin x$. The numerator starts telescoping: $2\sin x\cos x = \sin 2x$.

Step 2 — Keep doubling. $2\sin 2x\cos 2x = \sin 4x$; $2\sin 4x\cos 4x = \sin 8x$; $2\sin 8x\cos 8x = \sin 16x$.

Step 3 — Count the 2's. Four doublings introduce $2^4 = 16$ in the denominator: the product is $\frac{\sin 16x}{16\sin x}$. ∎

Takeaway: The doubling telescope in full generality: $\prod_{k=0}^{n-1}\cos 2^k x = \frac{\sin 2^n x}{2^n \sin x}$.

Example 31: From identity to value

Using $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$, evaluate $\sin^2 75° - \sin^2 15°$.

Solution:

Step 1 — Substitute A = 75°, B = 15°. $\sin^2 75° - \sin^2 15° = \sin 90°\sin 60°$.

Step 2 — Evaluate. $= 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.

Takeaway: One line via the identity versus messy surds by direct substitution — identities ARE the calculator.

Example 32: A final Board-style proof

Prove that $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$.

Solution:

Step 1 — Pair the outer terms around the middle. $\cos 4x + \cos 2x = 2\cos 3x\cos x$, so the numerator is $2\cos 3x\cos x + \cos 3x = \cos 3x(2\cos x + 1)$.

Step 2 — Same pairing below. $\sin 4x + \sin 2x = 2\sin 3x\cos x$, so the denominator is $\sin 3x(2\cos x + 1)$.

Step 3 — Cancel $(2\cos x + 1)$. Quotient $= \frac{\cos 3x}{\sin 3x} = \cot 3x$. ∎

Takeaway: Three-term sums: pair the OUTERS around the middle term; the middle's angle (3x) is always the surviving half-sum.