cot x = −5/√11, x∈(π/2,π) — Evaluate a Multi-Angle Trig Expression

JEE Advanced 2024, Paper 1. Previous Year Question.

Problem

Let $x\in\left(\dfrac{\pi}{2},\pi\right)$ be such that $\cot x = -\dfrac{5}{\sqrt{11}}$. Then

$$\left(\sin\frac{11x}{2}\right)\left(\sin6x-\cos6x\right) + \left(\cos\frac{11x}{2}\right)\left(\sin6x+\cos6x\right)$$

is equal to:

Key insight. Expanding the given expression term by term and regrouping reveals two hidden angle-sum/difference identities — the whole unwieldy expression collapses to just $\cos(x/2)+\sin(x/2)$. Squaring that sum turns it into $1+\sin x$, which is easy to find once $\cot x$ pins down a right triangle in the correct quadrant.

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Approach

Expand the given expression and regroup terms to reveal the angle-sum and angle-difference formulas for cosine and sine, collapsing everything to $\cos(x/2)+\sin(x/2)$. Square this sum to relate it to $1+\sin x$, find $\sin x$ from the given $\cot x$ (using the correct quadrant), and simplify the resulting nested square root by recognising it as a perfect square.

Solution

Step 1 — Expand the given expression

$$\sin\frac{11x}{2}\sin6x – \sin\frac{11x}{2}\cos6x + \cos\frac{11x}{2}\sin6x + \cos\frac{11x}{2}\cos6x$$

Step 2 — Group into two recognisable identities

Grouping the cos-cos and sin-sin terms:

$$\cos\frac{11x}{2}\cos6x+\sin\frac{11x}{2}\sin6x = \cos\left(6x-\frac{11x}{2}\right) = \cos\left(\frac{x}{2}\right)$$

Grouping the remaining two terms:

$$\cos\frac{11x}{2}\sin6x – \sin\frac{11x}{2}\cos6x = \sin\left(6x-\frac{11x}{2}\right) = \sin\left(\frac{x}{2}\right)$$

Step 3 — Combine

$$\text{Expression} = \cos\frac{x}{2}+\sin\frac{x}{2}$$

Step 4 — Square this sum

$$\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2 = 1+2\sin\frac{x}{2}\cos\frac{x}{2} = 1+\sin x$$

Step 5 — Find sin x from cot x

Since $\cot x=-\dfrac{5}{\sqrt{11}}$ and $x$ is in the second quadrant ($\sin x>0$, $\cos x<0$): using a right triangle with opposite $=\sqrt{11}$, adjacent $=5$ (magnitude), hypotenuse $=\sqrt{11+25}=\sqrt{36}=6$:

$$\sin x = \frac{\sqrt{11}}{6}$$

Step 6 — Substitute into 1 + sin x

$$1+\sin x = 1+\frac{\sqrt{11}}{6} = \frac{6+\sqrt{11}}{6}$$

Step 7 — Simplify the square root

$$\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2 = \frac{6+\sqrt{11}}{6}$$

Multiplying numerator and denominator by $2$:

$$\frac{12+2\sqrt{11}}{12} = \frac{11+2\sqrt{11}+1}{12} = \frac{(\sqrt{11}+1)^2}{12}$$

Since $\dfrac{x}{2}\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$, both $\sin\dfrac{x}{2}$ and $\cos\dfrac{x}{2}$ are positive, so the square root is positive:

$$\cos\frac{x}{2}+\sin\frac{x}{2} = \sqrt{\frac{(\sqrt{11}+1)^2}{12}} = \frac{\sqrt{11}+1}{\sqrt{12}} = \frac{\sqrt{11}+1}{2\sqrt3}$$

Answer

Common mistakes

• Missing that the nested square root is secretly a perfect square. Trying to evaluate $\sqrt{\frac{6+\sqrt{11}}{6}}$ directly, without first scaling by $2$ to reveal the $(\sqrt{11}+1)^2$ pattern, leaves the answer in an unrecognisable form that doesn’t match any option.
• Using the wrong sign for $\cos x$ or $\sin x$. Since $x$ is in the second quadrant, $\cos x$ is negative but $\sin x$ is positive — getting this backwards would flip the sign inside the square root and produce a completely different result.

Practise next

• Let $y\in\left(0,\dfrac{\pi}{2}\right)$ be such that $\tan y=\dfrac{3}{\sqrt5}$. Evaluate $\left(\cos\dfrac{9y}{2}\right)\left(\cos4y+\sin4y\right)-\left(\sin\dfrac{9y}{2}\right)\left(\cos4y-\sin4y\right)$, using the same expand-and-regroup approach.