NCERT Class 11 Mathematics — Trigonometric Functions, Exercise 3.3. All 25 questions solved.

This is the longest exercise in the chapter, and almost all of it is proof. Four families of identity between them settle every question:

Compound angles

$$\cos(x \pm y) = \cos x\cos y \mp \sin x \sin y, \qquad \sin(x \pm y) = \sin x\cos y \pm \cos x \sin y$$

$$\tan(x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y}$$

Sum to product

$$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}, \qquad \sin A – \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$$

$$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}, \qquad \cos A – \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$

Two useful consequences

$$\sin^2 A – \sin^2 B = \sin(A+B)\sin(A-B), \qquad \cos^2 A – \cos^2 B = -\sin(A+B)\sin(A-B)$$

Double and triple angles

$$\sin 2x = 2\sin x \cos x, \quad \cos 2x = 2\cos^2 x – 1 = 1 – 2\sin^2 x, \quad \cos 3x = 4\cos^3 x – 3\cos x$$

Key insight. Nearly every proof here is decided by which way you group the terms. In questions 16 to 21 the pattern is always the same: convert the sums in the numerator and denominator to products, and a common factor cancels, leaving a single tangent or cotangent. The grouping to choose is the one that makes the half-sum of the two angles match on both lines — in question 21, pairing $\cos 4x$ with $\cos 2x$ (half-sum $3x$) rather than with $\cos 3x$.

Prove that:

Question 1

$$\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} – \tan^2 \frac{\pi}{4} = -\frac{1}{2}$$

Solution. Substitute the standard values $\sin\frac{\pi}{6} = \frac12$, $\cos\frac{\pi}{3} = \frac12$, $\tan\frac{\pi}{4} = 1$:

$$\left(\frac12\right)^2 + \left(\frac12\right)^2 – 1^2 = \frac14 + \frac14 – 1 = \frac12 – 1 = -\frac{1}{2}$$

Question 2

$$2\sin^2 \frac{\pi}{6} + \operatorname{cosec}^2 \frac{7\pi}{6}\cos^2 \frac{\pi}{3} = \frac{3}{2}$$

Solution. The only term needing care is $\operatorname{cosec}\frac{7\pi}{6}$. Since $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$ lies in the third quadrant,

$$\sin\frac{7\pi}{6} = -\sin\frac{\pi}{6} = -\frac12 \quad\Longrightarrow\quad \operatorname{cosec}\frac{7\pi}{6} = -2, \quad \operatorname{cosec}^2\frac{7\pi}{6} = 4$$

Substituting:

$$2\left(\frac12\right)^2 + 4\left(\frac12\right)^2 = 2 \cdot \frac14 + 4 \cdot \frac14 = \frac12 + 1 = \frac{3}{2}$$

Note that squaring removes the minus sign, so the third-quadrant sign does not survive into the answer — but it must still be got right, because a student who writes $\operatorname{cosec}\frac{7\pi}{6} = 2$ has reasoned wrongly even though the number lands correctly here.

Question 3

$$\cot^2 \frac{\pi}{6} + \operatorname{cosec} \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} = 6$$

Solution. $\cot\frac{\pi}{6} = \sqrt3$, so $\cot^2\frac{\pi}{6} = 3$.

$\frac{5\pi}{6} = \pi – \frac{\pi}{6}$ is in the second quadrant, where sine is positive:

$$\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac12 \quad\Longrightarrow\quad \operatorname{cosec}\frac{5\pi}{6} = 2$$

And $\tan\frac{\pi}{6} = \frac{1}{\sqrt3}$, so $3\tan^2\frac{\pi}{6} = 3 \cdot \frac13 = 1$.

$$3 + 2 + 1 = 6$$

Question 4

$$2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10$$

Solution. $\frac{3\pi}{4} = \pi – \frac{\pi}{4}$, so $\sin\frac{3\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt2}$ and $\sin^2\frac{3\pi}{4} = \frac12$.

$\cos^2\frac{\pi}{4} = \frac12$, and $\sec\frac{\pi}{3} = 2$ so $\sec^2\frac{\pi}{3} = 4$.

$$2\left(\frac12\right) + 2\left(\frac12\right) + 2(4) = 1 + 1 + 8 = 10$$

Question 5

Find the value of: (i) $\sin 75^\circ$    (ii) $\tan 15^\circ$

Solution. Neither angle is standard, but each is a sum or difference of two that are.

(i) $75^\circ = 45^\circ + 30^\circ$, so use $\sin(x+y) = \sin x\cos y + \cos x \sin y$:

$$\sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$$

$$= \frac{1}{\sqrt2} \cdot \frac{\sqrt3}{2} + \frac{1}{\sqrt2} \cdot \frac12 = \frac{\sqrt3}{2\sqrt2} + \frac{1}{2\sqrt2} = \frac{\sqrt3 + 1}{2\sqrt2}$$

(ii) $15^\circ = 45^\circ – 30^\circ$, so use $\tan(x-y) = \dfrac{\tan x – \tan y}{1 + \tan x \tan y}$:

$$\tan 15^\circ = \frac{\tan 45^\circ – \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 – \frac{1}{\sqrt3}}{1 + \frac{1}{\sqrt3}} = \frac{\sqrt3 – 1}{\sqrt3 + 1}$$

Rationalise by multiplying above and below by $\sqrt3 – 1$:

$$= \frac{(\sqrt3-1)^2}{(\sqrt3)^2 – 1^2} = \frac{3 – 2\sqrt3 + 1}{2} = \frac{4 – 2\sqrt3}{2} = 2 – \sqrt3$$

Prove the following:

Question 6

$$\cos\left(\frac{\pi}{4} – x\right)\cos\left(\frac{\pi}{4} – y\right) – \sin\left(\frac{\pi}{4} – x\right)\sin\left(\frac{\pi}{4} – y\right) = \sin(x+y)$$

Solution. The left side has exactly the shape $\cos A \cos B – \sin A \sin B = \cos(A + B)$, with $A = \frac{\pi}{4} – x$ and $B = \frac{\pi}{4} – y$. There is no need to expand anything:

$$\text{LHS} = \cos\left[\left(\frac{\pi}{4} – x\right) + \left(\frac{\pi}{4} – y\right)\right] = \cos\left(\frac{\pi}{2} – x – y\right)$$

Now use $\cos\left(\frac{\pi}{2} – \theta\right) = \sin\theta$ with $\theta = x + y$:

$$= \sin(x + y) = \text{RHS} \qquad \blacksquare$$

Question 7

$$\frac{\tan\left(\frac{\pi}{4} + x\right)}{\tan\left(\frac{\pi}{4} – x\right)} = \left(\frac{1 + \tan x}{1 – \tan x}\right)^2$$

Solution. Expand each tangent using the compound-angle formula, remembering $\tan\frac{\pi}{4} = 1$:

$$\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 – \tan x}, \qquad \tan\left(\frac{\pi}{4} – x\right) = \frac{1 – \tan x}{1 + \tan x}$$

The two are reciprocals of each other, so dividing gives a square:

$$\frac{\tan\left(\frac{\pi}{4}+x\right)}{\tan\left(\frac{\pi}{4}-x\right)} = \frac{1 + \tan x}{1 – \tan x} \times \frac{1 + \tan x}{1 – \tan x} = \left(\frac{1 + \tan x}{1 – \tan x}\right)^2 \qquad \blacksquare$$

Question 8

$$\frac{\cos(\pi + x)\cos(-x)}{\sin(\pi – x)\cos\left(\frac{\pi}{2} + x\right)} = \cot^2 x$$

Solution. Reduce each of the four factors using the standard allied-angle results, taking care with signs:

$$\cos(\pi + x) = -\cos x, \qquad \cos(-x) = \cos x$$ $$\sin(\pi – x) = \sin x, \qquad \cos\left(\frac{\pi}{2} + x\right) = -\sin x$$

Substituting:

$$\text{LHS} = \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x \qquad \blacksquare$$

The two minus signs cancel, which is why the result is $+\cot^2 x$ — but both must be found, since losing exactly one flips the sign of the answer.

Question 9

$$\cos\left(\frac{3\pi}{2} + x\right)\cos(2\pi + x)\left[\cot\left(\frac{3\pi}{2} – x\right) + \cot(2\pi + x)\right] = 1$$

Solution. Reduce each allied angle first:

$$\cos\left(\frac{3\pi}{2} + x\right) = \sin x, \qquad \cos(2\pi + x) = \cos x$$ $$\cot\left(\frac{3\pi}{2} – x\right) = \tan x, \qquad \cot(2\pi + x) = \cot x$$

So

$$\text{LHS} = \sin x \cos x\,(\tan x + \cot x)$$

Now combine the bracket over a common denominator:

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

Therefore

$$\text{LHS} = \sin x \cos x \times \frac{1}{\sin x\cos x} = 1 \qquad \blacksquare$$

Question 10

$$\sin(n+1)x\,\sin(n+2)x + \cos(n+1)x\,\cos(n+2)x = \cos x$$

Solution. The left side is $\cos A\cos B + \sin A \sin B = \cos(A – B)$ with $A = (n+2)x$ and $B = (n+1)x$. Nothing needs expanding:

$$\text{LHS} = \cos\big[(n+2)x – (n+1)x\big] = \cos x \qquad \blacksquare$$

The parameter $n$ disappears entirely, because only the difference of the two angles survives — and that difference is $x$ whatever $n$ is.

Question 11

$$\cos\left(\frac{3\pi}{4} + x\right) – \cos\left(\frac{3\pi}{4} – x\right) = -\sqrt2 \sin x$$

Solution. Apply the difference-to-product formula $\cos A – \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$ with $A = \frac{3\pi}{4} + x$ and $B = \frac{3\pi}{4} – x$:

$$\frac{A+B}{2} = \frac{3\pi}{4}, \qquad \frac{A-B}{2} = x$$

$$\text{LHS} = -2\sin\frac{3\pi}{4}\sin x = -2 \cdot \frac{1}{\sqrt2} \cdot \sin x = -\sqrt2 \sin x \qquad \blacksquare$$

Question 12

$$\sin^2 6x – \sin^2 4x = \sin 2x \sin 10x$$

Solution. Use $\sin^2 A – \sin^2 B = \sin(A+B)\sin(A-B)$, which follows from the difference of two squares together with the product formulae. With $A = 6x$, $B = 4x$:

$$\text{LHS} = \sin(6x + 4x)\sin(6x – 4x) = \sin 10x \sin 2x \qquad \blacksquare$$

If the identity is not remembered, derive it in one line: $\sin^2 A – \sin^2 B = (\sin A + \sin B)(\sin A – \sin B)$, then convert each bracket to a product and simplify with $2\sin\theta\cos\theta = \sin 2\theta$.

Question 13

$$\cos^2 2x – \cos^2 6x = \sin 4x \sin 8x$$

Solution. The companion identity is $\cos^2 A – \cos^2 B = -\sin(A+B)\sin(A-B)$ — note the leading minus, which is the whole difference from question 12. With $A = 2x$, $B = 6x$:

$$\text{LHS} = -\sin(2x + 6x)\sin(2x – 6x) = -\sin 8x \sin(-4x)$$

Since $\sin(-\theta) = -\sin\theta$:

$$= -\sin 8x \cdot (-\sin 4x) = \sin 4x \sin 8x \qquad \blacksquare$$

Two minus signs cancel here — one from the identity and one from the negative angle.

Question 14

$$\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x$$

Solution. Group the outer two terms, whose half-sum is $4x$ — matching the $\sin 4x$ on the right:

$$\sin 2x + \sin 6x = 2\sin\frac{2x + 6x}{2}\cos\frac{6x – 2x}{2} = 2\sin 4x \cos 2x$$

So

$$\text{LHS} = 2\sin 4x\cos 2x + 2\sin 4x = 2\sin 4x\,(\cos 2x + 1)$$

Now use $\cos 2x = 2\cos^2 x – 1$, so $\cos 2x + 1 = 2\cos^2 x$:

$$\text{LHS} = 2\sin 4x \cdot 2\cos^2 x = 4\cos^2 x \sin 4x \qquad \blacksquare$$

Question 15

$$\cot 4x\,(\sin 5x + \sin 3x) = \cot x\,(\sin 5x – \sin 3x)$$

Solution. Convert both brackets to products. The half-sum is $4x$ either way, and that is what makes the identity work:

$$\sin 5x + \sin 3x = 2\sin 4x \cos x, \qquad \sin 5x – \sin 3x = 2\cos 4x \sin x$$

Now each side becomes the same expression:

$$\text{LHS} = \frac{\cos 4x}{\sin 4x} \cdot 2\sin 4x\cos x = 2\cos 4x\cos x$$

$$\text{RHS} = \frac{\cos x}{\sin x} \cdot 2\cos 4x \sin x = 2\cos 4x \cos x$$

$$\text{LHS} = \text{RHS} \qquad \blacksquare$$

The cotangent on each side is chosen precisely to cancel the sine that its own bracket produces — a neat piece of design worth noticing.

Question 16

$$\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x} = -\frac{\sin 2x}{\cos 10x}$$

Solution. Convert numerator and denominator to products.

$$\cos 9x – \cos 5x = -2\sin\frac{9x+5x}{2}\sin\frac{9x-5x}{2} = -2\sin 7x \sin 2x$$

$$\sin 17x – \sin 3x = 2\cos\frac{17x+3x}{2}\sin\frac{17x-3x}{2} = 2\cos 10x \sin 7x$$

Dividing, the factor $2\sin 7x$ cancels:

$$\text{LHS} = \frac{-2\sin 7x\sin 2x}{2\cos 10x \sin 7x} = -\frac{\sin 2x}{\cos 10x} \qquad \blacksquare$$

The unusual pairing of $9x$ with $5x$ and $17x$ with $3x$ is exactly what produces the common factor $\sin 7x$.

Question 17

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

Solution.

$$\sin 5x + \sin 3x = 2\sin 4x\cos x, \qquad \cos 5x + \cos 3x = 2\cos 4x \cos x$$

Both share the factor $2\cos x$:

$$\text{LHS} = \frac{2\sin 4x\cos x}{2\cos 4x\cos x} = \frac{\sin 4x}{\cos 4x} = \tan 4x \qquad \blacksquare$$

This is the template for questions 18 to 21: sum-to-product above and below, cancel the shared factor, and a single tangent (or cotangent) remains.

Question 18

$$\frac{\sin x – \sin y}{\cos x + \cos y} = \tan\frac{x – y}{2}$$

Solution.

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

The factor $2\cos\frac{x+y}{2}$ is common:

$$\text{LHS} = \frac{2\cos\frac{x+y}{2}\sin\frac{x-y}{2}}{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}} = \frac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}} = \tan\frac{x-y}{2} \qquad \blacksquare$$

Note that the half-sum cancels and the half-difference survives — the opposite of what happens in question 17, and it is the choice of $-$ in the numerator that decides which.

Question 19

$$\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$$

Solution. This is question 18’s pattern with $y$ replaced by $3x$ and both signs positive:

$$\sin x + \sin 3x = 2\sin 2x\cos x, \qquad \cos x + \cos 3x = 2\cos 2x\cos x$$

$$\text{LHS} = \frac{2\sin 2x\cos x}{2\cos 2x\cos x} = \tan 2x \qquad \blacksquare$$

Question 20

$$\frac{\sin x – \sin 3x}{\sin^2 x – \cos^2 x} = 2\sin x$$

Solution. Numerator, by the difference formula:

$$\sin x – \sin 3x = 2\cos\frac{x + 3x}{2}\sin\frac{x – 3x}{2} = 2\cos 2x \sin(-x) = -2\cos 2x\sin x$$

Denominator, using $\cos 2x = \cos^2 x – \sin^2 x$:

$$\sin^2 x – \cos^2 x = -(\cos^2 x – \sin^2 x) = -\cos 2x$$

Dividing:

$$\text{LHS} = \frac{-2\cos 2x\sin x}{-\cos 2x} = 2\sin x \qquad \blacksquare$$

Question 21

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

Solution. Three terms, so pair the outer two — their half-sum is $3x$, which matches the middle term and produces the common factor.

$$\cos 4x + \cos 2x = 2\cos 3x\cos x \quad\Longrightarrow\quad \text{numerator} = 2\cos 3x\cos x + \cos 3x = \cos 3x\,(2\cos x + 1)$$

$$\sin 4x + \sin 2x = 2\sin 3x\cos x \quad\Longrightarrow\quad \text{denominator} = 2\sin 3x\cos x + \sin 3x = \sin 3x\,(2\cos x + 1)$$

The bracket $(2\cos x + 1)$ is identical on both lines and cancels:

$$\text{LHS} = \frac{\cos 3x}{\sin 3x} = \cot 3x \qquad \blacksquare$$

Pairing $\cos 4x$ with $\cos 3x$ instead would give a half-sum of $\tfrac{7x}{2}$ and lead nowhere. Choosing the pairing that matches the leftover term is the whole technique.

Question 22

$$\cot x \cot 2x – \cot 2x \cot 3x – \cot 3x \cot x = 1$$

Solution. The key is to write $3x = 2x + x$ and use the cotangent addition formula:

$$\cot(A + B) = \frac{\cot A \cot B – 1}{\cot A + \cot B}$$

With $A = 2x$ and $B = x$:

$$\cot 3x = \frac{\cot 2x\cot x – 1}{\cot 2x + \cot x}$$

Cross-multiplying:

$$\cot 3x\,(\cot 2x + \cot x) = \cot 2x \cot x – 1$$

$$\cot 3x\cot 2x + \cot 3x\cot x = \cot x\cot 2x – 1$$

Rearranging so that all three products are on one side:

$$\cot x\cot 2x – \cot 2x\cot 3x – \cot 3x\cot x = 1 \qquad \blacksquare$$

Question 23

$$\tan 4x = \frac{4\tan x\,(1 – \tan^2 x)}{1 – 6\tan^2 x + \tan^4 x}$$

Solution. Write $t = \tan x$ and apply the double-angle formula twice.

First,

$$\tan 2x = \frac{2t}{1 – t^2}$$

Then

$$\tan 4x = \frac{2\tan 2x}{1 – \tan^2 2x}$$

Compute the two parts separately. The numerator:

$$2\tan 2x = \frac{4t}{1 – t^2}$$

The denominator, over the common denominator $(1-t^2)^2$:

$$1 – \tan^2 2x = 1 – \frac{4t^2}{(1-t^2)^2} = \frac{(1-t^2)^2 – 4t^2}{(1-t^2)^2} = \frac{1 – 2t^2 + t^4 – 4t^2}{(1-t^2)^2} = \frac{1 – 6t^2 + t^4}{(1-t^2)^2}$$

Dividing one by the other, one factor of $(1 – t^2)$ survives:

$$\tan 4x = \frac{4t}{1-t^2} \times \frac{(1-t^2)^2}{1 – 6t^2 + t^4} = \frac{4t\,(1 – t^2)}{1 – 6t^2 + t^4} \qquad \blacksquare$$

Question 24

$$\cos 4x = 1 – 8\sin^2 x\cos^2 x$$

Solution. Treat $4x$ as $2(2x)$ and use $\cos 2\theta = 1 – 2\sin^2\theta$ with $\theta = 2x$:

$$\cos 4x = 1 – 2\sin^2 2x$$

Now substitute $\sin 2x = 2\sin x\cos x$:

$$= 1 – 2\,(2\sin x\cos x)^2 = 1 – 2 \cdot 4\sin^2 x\cos^2 x = 1 – 8\sin^2 x\cos^2 x \qquad \blacksquare$$

Question 25

$$\cos 6x = 32\cos^6 x – 48\cos^4 x + 18\cos^2 x – 1$$

Solution. Treat $6x$ as $3(2x)$ and use the triple-angle formula $\cos 3\theta = 4\cos^3\theta – 3\cos\theta$ with $\theta = 2x$:

$$\cos 6x = 4\cos^3 2x – 3\cos 2x$$

Write $c = \cos x$, so $\cos 2x = 2c^2 – 1$. Expanding the cube:

$$(2c^2 – 1)^3 = 8c^6 – 3(4c^4)(1) + 3(2c^2)(1) – 1 = 8c^6 – 12c^4 + 6c^2 – 1$$

Therefore

$$4\cos^3 2x = 32c^6 – 48c^4 + 24c^2 – 4$$

$$-3\cos 2x = -3(2c^2 – 1) = -6c^2 + 3$$

Adding the two lines:

$$\cos 6x = 32c^6 – 48c^4 + (24 – 6)c^2 + (-4 + 3) = 32\cos^6 x – 48\cos^4 x + 18\cos^2 x – 1 \qquad \blacksquare$$

Going instead via $\cos 6x = 2\cos^2 3x – 1$ works equally well and is a good check: $2(4c^3 – 3c)^2 – 1$ expands to the same polynomial.

Common mistakes

• Question 2, taking $\operatorname{cosec}\frac{7\pi}{6}$ as positive. $\frac{7\pi}{6}$ is in the third quadrant, where sine is negative. The square hides the error in the final number but not in the reasoning.
• Questions 6 and 10, expanding everything. Both left sides already match a compound-angle formula exactly. Expanding $\cos(\frac{\pi}{4}-x)$ into $\cos\frac{\pi}{4}\cos x + \ldots$ turns a two-line proof into a page.
• Question 8, losing one of the two minus signs. $\cos(\pi + x) = -\cos x$ and $\cos(\frac{\pi}{2}+x) = -\sin x$; keeping one and dropping the other gives $-\cot^2 x$.
• Question 13, forgetting the minus in $\cos^2 A – \cos^2 B$. The sine version has no leading minus and the cosine version does. Getting them the same way round is the single most common slip in questions 12 and 13.
• Questions 14 and 21, pairing adjacent terms. In a three-term sum the outer two are the ones to combine, because their half-sum equals the middle angle and produces the common factor. Pairing neighbours leads to half-integer multiples of $x$ and a dead end.
• Question 16, pairing the wrong terms. $\cos 9x$ with $\cos 5x$ and $\sin 17x$ with $\sin 3x$ are the pairings given; both yield $7x$ as an argument, which is what cancels. There is nothing to choose here — but the arithmetic $\frac{17x+3x}{2} = 10x$ and $\frac{17x-3x}{2} = 7x$ must be done carefully.
• Question 20, mis-signing the denominator. $\sin^2 x – \cos^2 x = -\cos 2x$, not $\cos 2x$. Since the numerator also carries a minus, both are needed for the answer to come out positive.
• Question 23, applying the double-angle formula to $\tan x$ directly. $\tan 4x$ is $\tan(2 \cdot 2x)$, so the formula is applied twice, in terms of $\tan 2x$ and then $\tan x$. Trying to jump straight from $\tan x$ to $\tan 4x$ has no formula behind it.

Practise next

• Miscellaneous Exercise on Chapter 3 — more product-to-sum proofs, plus half-angle values $\sin\frac{x}{2}$, $\cos\frac{x}{2}$, $\tan\frac{x}{2}$ where the quadrant reasoning of Exercise 3.2 returns.
• Exercise 3.2 — worth revising alongside questions 1 to 4, since every allied angle there has to be reduced before its value can be written down.