Multiple Angles and Product-Sum Transformations

Double Angles — Set y = x and Harvest

Putting y = x in the sum formulas instantly produces the double-angle formulas:

$$\sin 2x = 2\sin x \cos x$$ $$\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$$ $$\tan 2x = \frac{2\tan x}{1 - \tan^2 x} \quad \left(2x \neq n\pi + \frac{\pi}{2}\right)$$

The three faces of cos 2x come from substituting $\sin^2 x = 1 - \cos^2 x$ or $\cos^2 x = 1 - \sin^2 x$ — pick whichever face matches the problem.

There are also tan-only forms (divide through by $\cos^2 x$):

$$\sin 2x = \frac{2\tan x}{1 + \tan^2 x} \qquad \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$$

Power reduction — reading cos 2x backwards

$$\cos^2 x = \frac{1 + \cos 2x}{2} \qquad \sin^2 x = \frac{1 - \cos 2x}{2}$$

These turn squares into first powers — the key to half-angle values like $\cos\frac{\pi}{8}$ and, later, to integration.

[JEE Tip] $1 + \cos 2x = 2\cos^2 x$ and $1 - \cos 2x = 2\sin^2 x$ — recognising "1 ± cos(something)" as a perfect double-angle square is the single most-used move in JEE trig simplification. Likewise $1 \pm \sin 2x = (\sin x \pm \cos x)^2$.

Triple Angles

Writing 3x = 2x + x and expanding with the sum and double-angle formulas:

$$\sin 3x = 3\sin x - 4\sin^3 x$$ $$\cos 3x = 4\cos^3 x - 3\cos x$$ $$\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \quad \left(3x \neq n\pi + \frac{\pi}{2}\right)$$

Memory anchors: sin 3x starts with 3 sin x (and subtracts the cube term); cos 3x starts with the cube term (and subtracts 3 cos x) — they are mirror-shaped.

Derivation sketch for sin 3x (examinable): $\sin(2x + x) = \sin 2x\cos x + \cos 2x \sin x = 2\sin x\cos^2 x + (1 - 2\sin^2 x)\sin x = 2\sin x(1 - \sin^2 x) + \sin x - 2\sin^3 x = 3\sin x - 4\sin^3 x$. ∎

Key Point: These formulas run BOTH ways: $4\cos^3 x = \cos 3x + 3\cos x$ lets you reduce cubes, and the factored forms $\sin 3x = \sin x(3 - 4\sin^2 x) = \sin x (2\cos 2x + 1)$ appear in product identities.

[JEE Tip] From cos 3x with $x = \frac{\pi}{9}$ (20°): $4\cos^3 20° - 3\cos 20° = \cos 60° = \frac{1}{2}$ — the reason $\cos 20°$ satisfies the cubic $8t^3 - 6t - 1 = 0$. JEE loves this bridge between trigonometry and cubic equations.

Product ↔ Sum Transformations

Adding and subtracting the four sum-difference formulas yields two interchangeable toolkits:

Products to sums

$$2\cos x \cos y = \cos(x + y) + \cos(x - y)$$ $$-2\sin x \sin y = \cos(x + y) - \cos(x - y)$$ $$2\sin x \cos y = \sin(x + y) + \sin(x - y)$$ $$2\cos x \sin y = \sin(x + y) - \sin(x - y)$$

Sums to products (substitute $\theta = x + y$, $\phi = x - y$)

$$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2} \qquad \cos x - \cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2}$$ $$\sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2} \qquad \sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2}$$

When to use which: sums → products when you want to FACTOR (prove quotient identities, solve equations); products → sums when you want to SPLIT (evaluate products of cosines, prepare for integration).

Key Point (sign traps): $\cos x - \cos y$ opens with $-2\sin\ldots\sin$ — the only formula with a leading minus. And in $\sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2}$, the HALF-DIFFERENCE sits inside the sine.

[Board Important] Identity questions of the type $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$ are near-guaranteed: convert both sums to products, cancel the common factor, read off the tangent.

Solved Examples

Example 1: Double-angle values from tan x

If $\tan x = \frac{3}{4}$ with x acute, find sin 2x, cos 2x and tan 2x.

Solution:

Step 1 — Use the tan-only form for sin 2x. $\sin 2x = \frac{2\tan x}{1 + \tan^2 x} = \frac{2 \cdot \frac{3}{4}}{1 + \frac{9}{16}} = \frac{3/2}{25/16} = \frac{24}{25}$.

Step 2 — Tan-only form for cos 2x. $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} = \frac{7/16}{25/16} = \frac{7}{25}$.

Step 3 — Tan-only form for tan 2x. $\tan 2x = \frac{2\tan x}{1 - \tan^2 x} = \frac{3/2}{7/16} = \frac{24}{7}$.

Step 4 — Cross-check. $\frac{\sin 2x}{\cos 2x} = \frac{24/25}{7/25} = \frac{24}{7} = \tan 2x$. ✓

Takeaway: The tan-only forms skip finding sin x and cos x entirely — fastest route when tan is given.

Example 2: Half-angle value via power reduction

Find the exact value of $\cos\frac{\pi}{8}$.

Solution:

Step 1 — Power-reduce with $x = \frac{\pi}{8}$. $\cos^2\frac{\pi}{8} = \frac{1 + \cos\frac{\pi}{4}}{2} = \frac{1 + \frac{1}{\sqrt{2}}}{2}$.

Step 2 — Tidy the fraction. Multiply top and bottom by $\sqrt{2}$: $\frac{\sqrt{2} + 1}{2\sqrt{2}} = \frac{2 + \sqrt{2}}{4}$.

Step 3 — Take the quadrant-decided root. $\frac{\pi}{8}$ is a first-quadrant angle, so cosine is positive: $\cos\frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$.

Takeaway: Halving angles = power reduction + a quadrant-decided square root. The nested surd is the expected form.

Example 3: Proving a 1 ± cos identity

Prove that $\frac{\sin 2x}{1 + \cos 2x} = \tan x$, and use it to evaluate tan 22.5°.

Solution:

Step 1 — Replace numerator and denominator by their squares. $\sin 2x = 2\sin x \cos x$ and $1 + \cos 2x = 2\cos^2 x$.

Step 2 — Cancel. $\frac{2\sin x \cos x}{2\cos^2 x} = \frac{\sin x}{\cos x} = \tan x$. ∎

Step 3 — Apply with 2x = 45°. $\tan 22.5° = \frac{\sin 45°}{1 + \cos 45°} = \frac{\frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2} + 1}$.

Step 4 — Rationalise. Multiply by $\frac{\sqrt{2} - 1}{\sqrt{2} - 1}$: $\tan 22.5° = \sqrt{2} - 1$.

Takeaway: The half-angle-tangent identity $\tan x = \frac{\sin 2x}{1 + \cos 2x}$ is a value factory: 22.5°, 15°, 75° all fall to it.

Example 4: A sin 3x factoring

Prove that $\sin 3x + \sin x = 2\sin 2x \cos x = 4\sin x \cos^2 x$.

Solution:

Step 1 — Sum to product. $\sin 3x + \sin x = 2\sin\frac{3x + x}{2}\cos\frac{3x - x}{2} = 2\sin 2x \cos x$ — first target reached.

Step 2 — Expand the double angle. $2\sin 2x\cos x = 2(2\sin x\cos x)\cos x = 4\sin x \cos^2 x$ — second target. ∎

Takeaway: Both target forms come from the SAME factoring — stop at whichever the question asks for.

Example 5: The tan 4x quotient

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

Solution:

Step 1 — Factor the numerator. $\sin 5x + \sin 3x = 2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2} = 2\sin 4x \cos x$.

Step 2 — Factor the denominator. $\cos 5x + \cos 3x = 2\cos 4x \cos x$.

Step 3 — Cancel the common factor. $\frac{2\sin 4x\cos x}{2\cos 4x\cos x} = \frac{\sin 4x}{\cos 4x} = \tan 4x$. ∎

Takeaway: Matching half-sums (4x) and half-differences (x) top and bottom guarantee a clean cancellation — this is the flagship sum-to-product pattern.

Example 6: A minus-sign quotient

Prove that $\frac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \cot x$.

Solution:

Step 1 — Factor the numerator. $\cos 7x + \cos 5x = 2\cos 6x \cos x$.

Step 2 — Factor the denominator, minding the minus. $\sin 7x - \sin 5x = 2\cos\frac{7x+5x}{2}\sin\frac{7x-5x}{2} = 2\cos 6x \sin x$.

Step 3 — Cancel $2\cos 6x$. $\frac{\cos x}{\sin x} = \cot x$. ∎

Takeaway: $\sin A - \sin B$ produces a COSINE of the half-sum — exactly what cancels against the numerator. Trust the formulas' asymmetry.

Example 7: Product of three cosines

Prove that $\cos 20° \cos 40° \cos 80° = \frac{1}{8}$.

Solution:

Step 1 — Multiply and divide by $2\sin 20°$. The product becomes $\frac{(2\sin 20°\cos 20°)\cos 40° \cos 80°}{2\sin 20°} = \frac{\sin 40°\cos 40°\cos 80°}{2\sin 20°}$.

Step 2 — Telescope again. $\sin 40°\cos 40° = \frac{1}{2}\sin 80°$, so the expression is $\frac{\sin 80°\cos 80°}{4\sin 20°} = \frac{\sin 160°}{8\sin 20°}$.

Step 3 — Reduce the top. $\sin 160° = \sin(180° - 20°) = \sin 20°$, which cancels: the product is $\frac{1}{8}$. ∎

Takeaway: Angles doubling in a cosine product (20°, 40°, 80°) invite the $\frac{\sin 2\theta}{2\sin\theta}$ telescope — a JEE evergreen.

Example 8: cos 2x from cos x, both directions

If $\cos x = -\frac{3}{5}$ with x in Q II, find cos 2x and sin 2x, and locate the quadrant of 2x.

Solution:

Step 1 — cos 2x needs only cos x. $\cos 2x = 2\cos^2 x - 1 = 2 \cdot \frac{9}{25} - 1 = \frac{18 - 25}{25} = -\frac{7}{25}$.

Step 2 — Complete sin x with the quadrant sign. $\sin^2 x = 1 - \frac{9}{25} = \frac{16}{25}$, and Q II keeps sine positive: $\sin x = \frac{4}{5}$.

Step 3 — sin 2x. $\sin 2x = 2\sin x\cos x = 2 \cdot \frac{4}{5} \cdot \left(-\frac{3}{5}\right) = -\frac{24}{25}$.

Step 4 — Read the quadrant of 2x from the signs. cos 2x < 0 and sin 2x < 0: Q III. Consistent: $x \in \left(\frac{\pi}{2}, \pi\right)$ gives $2x \in (\pi, 2\pi)$. ✓

Takeaway: Doubling the angle can land in any quadrant — compute both values and read the signs rather than guessing.

Example 9: Proving with power reduction

Prove that $\cos^2 x + \cos^2\left(x + \frac{\pi}{3}\right) + \cos^2\left(x - \frac{\pi}{3}\right) = \frac{3}{2}$.

Solution:

Step 1 — Power-reduce every square. Each $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$, so the sum is $\frac{3}{2} + \frac{1}{2}\left[\cos 2x + \cos\left(2x + \frac{2\pi}{3}\right) + \cos\left(2x - \frac{2\pi}{3}\right)\right]$.

Step 2 — Sum-to-product on the shifted pair. $\cos\left(2x + \frac{2\pi}{3}\right) + \cos\left(2x - \frac{2\pi}{3}\right) = 2\cos 2x \cos\frac{2\pi}{3} = 2\cos 2x \cdot \left(-\frac{1}{2}\right) = -\cos 2x$.

Step 3 — Watch the bracket collapse. $\cos 2x - \cos 2x = 0$, leaving exactly $\frac{3}{2}$ — independent of x. ∎

Takeaway: Sums of squared cosines at symmetric shifts: power-reduce, then let sum-to-product annihilate the oscillating part.

Example 10: sin 3x in action

If $\sin x = \frac{1}{3}$, find the exact value of sin 3x.

Solution:

Step 1 — Quote the triple-angle formula. $\sin 3x = 3\sin x - 4\sin^3 x$ — a polynomial in sin x alone.

Step 2 — Substitute. $= 3 \cdot \frac{1}{3} - 4 \cdot \left(\frac{1}{3}\right)^3 = 1 - \frac{4}{27}$.

Step 3 — Combine. $= \frac{27 - 4}{27} = \frac{23}{27}$.

Takeaway: No need for cos x at all — sin 3x is a polynomial in sin x alone (and cos 3x in cos x alone).

Example 11: The famous cot-tan collapse

Prove that $\cot x - \tan x = 2\cot 2x$.

Solution:

Step 1 — Put both terms over one denominator. $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x}$.

Step 2 — Recognise double angles top and bottom. Numerator $= \cos 2x$; denominator $\sin x\cos x = \frac{1}{2}\sin 2x$.

Step 3 — Divide. $\frac{\cos 2x}{\frac{1}{2}\sin 2x} = \frac{2\cos 2x}{\sin 2x} = 2\cot 2x$. ∎

Takeaway: $\sin x\cos x = \frac{1}{2}\sin 2x$ upgrades denominators instantly; its family: $\tan x + \cot x = \frac{2}{\sin 2x}$.

Example 12: A product-to-sum evaluation

Evaluate $2\sin 75° \sin 15°$, and hence $\sin 75° \sin 15°$.

Solution:

Step 1 — Apply product-to-sum. $2\sin x \sin y = \cos(x - y) - \cos(x + y)$ with x = 75°, y = 15°.

Step 2 — Substitute the angles. $= \cos 60° - \cos 90° = \frac{1}{2} - 0 = \frac{1}{2}$.

Step 3 — Halve for the plain product. $\sin 75°\sin 15° = \frac{1}{4}$.

Takeaway: Products of sines/cosines at complementary or supplementary pairs collapse to standard values via the transformation formulas — no surds needed.

Example 13: A four-term sum to product

Prove that $\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$, and then that $(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x = 0$.

Solution:

Step 1 — First identity: factor top and bottom. Numerator $= 2\sin 2x\cos x$; denominator $= 2\cos 2x \cos x$; cancelling $2\cos x$ leaves $\tan 2x$. ∎

Step 2 — Second identity: expand the brackets. $\sin 3x\sin x + \sin^2 x + \cos 3x\cos x - \cos^2 x$.

Step 3 — Group into recognisable patterns. $(\cos 3x\cos x + \sin 3x \sin x) - (\cos^2 x - \sin^2 x) = \cos(3x - x) - \cos 2x$.

Step 4 — Collapse. $\cos 2x - \cos 2x = 0$. ∎

Takeaway: Two different toolkits (sum-to-product; difference formula recognised in reverse) — choose by SHAPE: quotients want factoring, mixed dot-product shapes want cos(A − B).

Example 14: cos 6x in terms of cos 2x

Express cos 6x as a polynomial in cos 2x, and verify at x = 30°.

Solution:

Step 1 — Nest the multiple angles. $6x = 3(2x)$, so apply the triple-angle formula to the angle 2x: $\cos 6x = 4\cos^3 2x - 3\cos 2x$.

Step 2 — Verify at x = 30°. LHS: $\cos 180° = -1$. RHS: $4\cos^3 60° - 3\cos 60° = 4 \cdot \frac{1}{8} - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = -1$. ✓

Takeaway: Multiple-angle formulas nest: 6x = 3(2x) = 2(3x) — pick the nesting that matches the target variable.