NCERT Class 11 Mathematics — Trigonometric Functions, Miscellaneous Exercise on Chapter 3. All 10 questions solved.

The exercise falls into two halves. Questions 1 to 7 are proofs, and every one of them is settled by converting sums into products; questions 8 to 10 ask for half-angle values, where the quadrant work of Exercise 3.2 returns with an extra twist — you must locate $\tfrac{x}{2}$, not $x$.

The formulae in play:

$$\cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}, \qquad \sin^2\frac{\theta}{2} = \frac{1 – \cos\theta}{2}$$

$$2\cos A\cos B = \cos(A+B) + \cos(A-B)$$

Key insight. For questions 8 to 10, halve the interval, not just the angle. If $x$ lies in quadrant II then $90^\circ < x < 180^\circ$, so $45^\circ < \tfrac{x}{2} < 90^\circ$ — quadrant I, where everything is positive. Quadrant III halves to quadrant II. Skipping this step leaves the sign of every answer to guesswork.

Prove that:

Question 1

$$2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0$$

Solution. Turn the product into a sum using $2\cos A\cos B = \cos(A+B) + \cos(A-B)$, with $A = \frac{\pi}{13}$ and $B = \frac{9\pi}{13}$:

$$2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} = \cos\frac{10\pi}{13} + \cos\left(-\frac{8\pi}{13}\right) = \cos\frac{10\pi}{13} + \cos\frac{8\pi}{13}$$

using $\cos(-\theta) = \cos\theta$. So the left side is

$$\cos\frac{10\pi}{13} + \cos\frac{8\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13}$$

Now note that both of the new angles are supplements of the two already present:

$$\frac{10\pi}{13} = \pi – \frac{3\pi}{13}, \qquad \frac{8\pi}{13} = \pi – \frac{5\pi}{13}$$

and $\cos(\pi – \theta) = -\cos\theta$, so

$$\cos\frac{10\pi}{13} = -\cos\frac{3\pi}{13}, \qquad \cos\frac{8\pi}{13} = -\cos\frac{5\pi}{13}$$

The four terms cancel in pairs:

$$-\cos\frac{3\pi}{13} – \cos\frac{5\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0 \qquad \blacksquare$$

The choice of $13$ is what makes the supplements line up; nothing about the number itself matters beyond that.

Question 2

$$(\sin 3x + \sin x)\sin x + (\cos 3x – \cos x)\cos x = 0$$

Solution. Expand the brackets and group by what the compound-angle formulae can absorb:

$$\text{LHS} = \sin 3x\sin x + \sin^2 x + \cos 3x\cos x – \cos^2 x$$

The first and third terms make $\cos(3x – x)$:

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

The second and fourth make $-(\cos^2 x – \sin^2 x) = -\cos 2x$. Therefore

$$\text{LHS} = \cos 2x – \cos 2x = 0 \qquad \blacksquare$$

Question 3

$$(\cos x + \cos y)^2 + (\sin x – \sin y)^2 = 4\cos^2\frac{x+y}{2}$$

Solution. Expand both squares:

$$\cos^2 x + 2\cos x\cos y + \cos^2 y + \sin^2 x – 2\sin x\sin y + \sin^2 y$$

Collect the Pythagorean pairs $\cos^2 x + \sin^2 x = 1$ and $\cos^2 y + \sin^2 y = 1$:

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

Now apply $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$ with $\theta = x + y$:

$$= 2\big(1 + \cos(x+y)\big) = 2 \cdot 2\cos^2\frac{x+y}{2} = 4\cos^2\frac{x+y}{2} \qquad \blacksquare$$

The minus sign inside the second bracket is what produces $\cos x\cos y – \sin x\sin y$ — that is, $\cos(x+y)$ rather than $\cos(x-y)$.

Question 4

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

Solution. Expanding as before:

$$\cos^2 x – 2\cos x\cos y + \cos^2 y + \sin^2 x – 2\sin x\sin y + \sin^2 y$$

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

Now use $1 – \cos\theta = 2\sin^2\frac{\theta}{2}$ with $\theta = x – y$:

$$= 2\big(1 – \cos(x-y)\big) = 4\sin^2\frac{x-y}{2} \qquad \blacksquare$$

Compare with question 3: changing one sign inside the brackets flips $\cos(x+y)$ to $\cos(x-y)$ and, with it, the whole answer from a cosine of the half-sum to a sine of the half-difference. (Geometrically, both sides here are the square of the distance between two points on the unit circle.)

Question 5

$$\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos 2x\sin 4x$$

Solution. Pair the terms so that both pairs share the same half-sum. Taking the outermost pair and the innermost pair, both give $4x$:

$$\sin x + \sin 7x = 2\sin 4x\cos 3x, \qquad \sin 3x + \sin 5x = 2\sin 4x\cos x$$

Adding and factoring out $2\sin 4x$:

$$\text{LHS} = 2\sin 4x\,(\cos 3x + \cos x)$$

Convert the remaining sum to a product:

$$\cos 3x + \cos x = 2\cos 2x\cos x$$

Therefore

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

Question 6

$$\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x$$

Solution. The brackets are already grouped for you, and every pair has half-sum $6x$ — that is the whole design of the question.

Numerator:

$$\sin 7x + \sin 5x = 2\sin 6x\cos x, \qquad \sin 9x + \sin 3x = 2\sin 6x\cos 3x$$

$$\text{numerator} = 2\sin 6x(\cos x + \cos 3x)$$

Denominator:

$$\cos 7x + \cos 5x = 2\cos 6x\cos x, \qquad \cos 9x + \cos 3x = 2\cos 6x\cos 3x$$

$$\text{denominator} = 2\cos 6x(\cos x + \cos 3x)$$

The factor $2(\cos x + \cos 3x)$ is common and cancels:

$$\text{LHS} = \frac{\sin 6x}{\cos 6x} = \tan 6x \qquad \blacksquare$$

Question 7

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

Solution. The right side contains $\sin x$ as a factor, so aim to produce it. Group the two terms that differ by $2x$ — namely $\sin 3x$ and $-\sin x$:

$$\sin 3x – \sin x = 2\cos 2x\sin x$$

So

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

Now convert the remaining sum:

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

Therefore

$$\text{LHS} = 2\sin x \cdot 2\cos\frac{3x}{2}\cos\frac{x}{2} = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2} \qquad \blacksquare$$

The half-angles on the right are the clue that the final step must combine two cosines whose arguments differ by an odd amount — here $2x$ and $x$, giving $\tfrac{3x}{2}$ and $\tfrac{x}{2}$.

Find $\sin\dfrac{x}{2}$, $\cos\dfrac{x}{2}$ and $\tan\dfrac{x}{2}$ in each of the following:

Question 8

$\tan x = -\dfrac{4}{3}$, $x$ in quadrant II

Solution. First get $\cos x$. In quadrant II, $\cos x < 0$, and from $\sec^2 x = 1 + \tan^2 x = 1 + \tfrac{16}{9} = \tfrac{25}{9}$:

$$\sec x = -\frac{5}{3} \quad\Longrightarrow\quad \cos x = -\frac{3}{5}$$

Now locate $\tfrac{x}{2}$. Since $x$ is in quadrant II,

$$\frac{\pi}{2} < x < \pi \quad\Longrightarrow\quad \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$$

so $\tfrac{x}{2}$ lies in quadrant I and all three values are positive.

$$\sin^2\frac{x}{2} = \frac{1 – \cos x}{2} = \frac{1 + \frac35}{2} = \frac{8/5}{2} = \frac{4}{5} \quad\Longrightarrow\quad \sin\frac{x}{2} = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}$$

$$\cos^2\frac{x}{2} = \frac{1 + \cos x}{2} = \frac{1 – \frac35}{2} = \frac{2/5}{2} = \frac{1}{5} \quad\Longrightarrow\quad \cos\frac{x}{2} = \frac{1}{\sqrt5} = \frac{\sqrt5}{5}$$

$$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{2/\sqrt5}{1/\sqrt5} = 2$$

A note on the printed answer. NCERT’s key gives the first two values as above but prints $\tan\frac{x}{2} = \frac12$. That contradicts its own first two entries, since $\frac{2\sqrt5/5}{\sqrt5/5} = 2$. Checking numerically: $\tan x = -\frac43$ in quadrant II puts $x = 126.8699^\circ$, so $\frac{x}{2} = 63.4349^\circ$ and $\tan\frac{x}{2} = 2.0000$. The printed $\frac12$ is the reciprocal of the correct value, and independent solution sources give $2$ as well. The answers to questions 9 and 10 are correct as printed.

Question 9

$\cos x = -\dfrac{1}{3}$, $x$ in quadrant III

Solution. Locate $\tfrac{x}{2}$ first. Since $x$ is in quadrant III,

$$\pi < x < \frac{3\pi}{2} \quad\Longrightarrow\quad \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$$

so $\tfrac{x}{2}$ lies in quadrant II: sine positive, cosine and tangent negative. This is the step that decides all three signs, and $x$ being in quadrant III does not mean $\tfrac{x}{2}$ is.

$$\sin^2\frac{x}{2} = \frac{1 – \cos x}{2} = \frac{1 + \frac13}{2} = \frac{4/3}{2} = \frac{2}{3} \quad\Longrightarrow\quad \sin\frac{x}{2} = +\sqrt{\frac23} = \frac{\sqrt6}{3}$$

$$\cos^2\frac{x}{2} = \frac{1 + \cos x}{2} = \frac{1 – \frac13}{2} = \frac{2/3}{2} = \frac{1}{3} \quad\Longrightarrow\quad \cos\frac{x}{2} = -\frac{1}{\sqrt3} = -\frac{\sqrt3}{3}$$

$$\tan\frac{x}{2} = \frac{\sqrt6/3}{-\sqrt3/3} = -\frac{\sqrt6}{\sqrt3} = -\sqrt2$$

Question 10

$\sin x = \dfrac{1}{4}$, $x$ in quadrant II

Solution. In quadrant II, $\cos x < 0$:

$$\cos x = -\sqrt{1 – \frac{1}{16}} = -\frac{\sqrt{15}}{4}$$

As in question 8, $x$ in quadrant II puts $\tfrac{x}{2}$ in quadrant I, so all three values are positive.

$$\sin^2\frac{x}{2} = \frac{1 – \cos x}{2} = \frac{1 + \frac{\sqrt{15}}{4}}{2} = \frac{4 + \sqrt{15}}{8}$$

Rationalise into the form NCERT prints by multiplying above and below by $2$:

$$\sin\frac{x}{2} = \sqrt{\frac{4+\sqrt{15}}{8}} = \sqrt{\frac{8 + 2\sqrt{15}}{16}} = \frac{\sqrt{8 + 2\sqrt{15}}}{4}$$

Similarly

$$\cos^2\frac{x}{2} = \frac{1 + \cos x}{2} = \frac{4 – \sqrt{15}}{8} \quad\Longrightarrow\quad \cos\frac{x}{2} = \frac{\sqrt{8 – 2\sqrt{15}}}{4}$$

For the tangent, take the ratio and rationalise:

$$\tan\frac{x}{2} = \sqrt{\frac{4 + \sqrt{15}}{4 – \sqrt{15}}} = \sqrt{\frac{(4+\sqrt{15})^2}{(4-\sqrt{15})(4+\sqrt{15})}} = \sqrt{\frac{(4+\sqrt{15})^2}{16 – 15}} = 4 + \sqrt{15}$$

The denominator collapses to $1$, which is why the answer is exact rather than another nested radical.

Common mistakes

• Question 1, expanding $\cos\frac{9\pi}{13}$ numerically. The proof is exact and needs no decimals: turn the product into a sum, then recognise both new angles as supplements of the two already present.
• Question 3 versus 4, mixing up which sign gives which. $(\cos x + \cos y)^2$ with $(\sin x – \sin y)^2$ produces $\cos(x+y)$ and a half-sum answer; two minus signs produce $\cos(x-y)$ and a half-difference answer.
• Questions 5 and 6, pairing adjacent terms. Pair so that every pair has the same half-sum. In question 5, $\sin x$ with $\sin 7x$ and $\sin 3x$ with $\sin 5x$ both give $4x$; pairing neighbours gives $2x$ and $6x$ and no common factor.
• Question 7, expanding $\sin 3x$ into $3\sin x – 4\sin^3 x$. It is not wrong, but it leads to a cubic that then has to be re-factored. Grouping $\sin 3x – \sin x$ reaches the answer in two lines.
• Questions 8 to 10, halving the angle but not the interval. $x$ in quadrant III gives $\tfrac{x}{2}$ in quadrant II, not quadrant III. Writing down the inequality for $x$ and dividing it by $2$ takes one line and fixes every sign.
• Question 9, making $\cos\frac{x}{2}$ positive. In quadrant II cosine is negative, so the minus must be inserted by hand after taking the square root — the formula $\cos^2\frac{x}{2} = \frac{1+\cos x}{2}$ gives only the magnitude.
• Question 10, leaving $\tan\frac{x}{2}$ as a nested radical. Rationalising $\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}$ makes the denominator $16 – 15 = 1$ and the answer collapses to $4 + \sqrt{15}$.

Practise next

• Exercise 4.1 — the next chapter, Complex Numbers and Quadratic Equations, where the modulus-argument form leans on exactly the standard angles used here.
• Exercise 3.3 — worth working again alongside questions 1 to 7, since the sum-to-product technique is identical and the practice compounds.