JEE Main 2019. Previous Year Question.
Problem
The value of $\cos\dfrac{\pi}{2^2}\cdot\cos\dfrac{\pi}{2^3}\cdots\cos\dfrac{\pi}{2^{10}}\cdot\sin\dfrac{\pi}{2^{10}}$ is:
Key insight. The identity $\sin\theta\cos\theta=\tfrac12\sin2\theta$ lets a $\sin$-times-$\cos$ pair collapse into a single $\sin$ of double the angle. Starting from the innermost $\sin(\pi/2^{10})\cos(\pi/2^{10})$ pair and working outward, each of the nine cosines gets absorbed one at a time, each absorption doubling the angle and halving the running total.
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Approach
Let $\theta=\pi/2^{10}$, so the product is $\sin\theta\cdot\cos\theta\cdot\cos2\theta\cdot\cos4\theta\cdots\cos(2^8\theta)$ (the cosine angles $\pi/2^9,\pi/2^8,\ldots,\pi/2^2$ are exactly $2\theta,4\theta,\ldots,2^8\theta$). Repeatedly apply $\sin x\cos x=\tfrac12\sin2x$, absorbing one cosine at a time, until only a single sine of a much larger angle remains.
Solution
Step 1 — Rewrite every angle in terms of θ = π/2¹⁰
$$\frac{\pi}{2^9}=2\theta,\quad \frac{\pi}{2^8}=4\theta,\quad \ldots,\quad \frac{\pi}{2^2}=2^8\theta$$
So the product is:
$$P = \sin\theta\cdot\cos\theta\cdot\cos(2\theta)\cdot\cos(4\theta)\cdots\cos(2^8\theta)$$
Step 2 — Absorb the first cosine
$$\sin\theta\cos\theta = \frac12\sin(2\theta)$$
$$P = \frac12\sin(2\theta)\cdot\cos(2\theta)\cdot\cos(4\theta)\cdots\cos(2^8\theta)$$
Step 3 — Absorb the second cosine
$$\sin(2\theta)\cos(2\theta) = \frac12\sin(4\theta)$$
$$P = \frac{1}{2^2}\sin(4\theta)\cdot\cos(4\theta)\cdots\cos(2^8\theta)$$
Step 4 — Continue the pattern
Each absorption doubles the sine’s angle and multiplies the running factor by $\tfrac12$. After absorbing all $9$ cosines (from $\cos\theta$ through $\cos(2^8\theta)$):
$$P = \frac{1}{2^9}\sin(2^9\theta)$$
Step 5 — Evaluate the final angle
$$2^9\theta = 512\cdot\frac{\pi}{1024} = \frac{\pi}{2}$$
$$P = \frac{1}{512}\sin\frac{\pi}{2} = \frac{1}{512}\cdot1 = \frac{1}{512}$$
Answer
$$\frac{1}{512}$$
Common mistakes
- Losing track of how many cosines have been absorbed. With nine cosine factors (from $2^2$ down to $2^{10}$ in the denominators), it’s easy to stop one step early or late — writing out the angle explicitly at each stage (as $2\theta,4\theta,8\theta,\ldots$) keeps the count honest.
- Trying to expand each cosine numerically instead of using the telescoping identity. $\cos(\pi/1024)$ and similar values have no simple closed form individually — the whole problem is designed to be solved via the $\sin\theta\cos\theta$ collapse, not by evaluating terms one at a time.
Practise next
- Find the value of $\cos\dfrac{\pi}{2^2}\cdot\cos\dfrac{\pi}{2^3}\cdots\cos\dfrac{\pi}{2^{8}}\cdot\sin\dfrac{\pi}{2^{8}}$, using the same telescoping approach with fewer terms.
Show answer
$\dfrac{1}{128}$. Start from the right: $\sin\theta\cos\theta=\tfrac12\sin2\theta$ with $\theta=\tfrac{\pi}{2^8}$.
Each step absorbs the next cosine and halves: $\sin\tfrac{\pi}{2^8}\cos\tfrac{\pi}{2^8}=\tfrac12\sin\tfrac{\pi}{2^7}$, then $\tfrac12\sin\tfrac{\pi}{2^7}\cos\tfrac{\pi}{2^7}=\tfrac14\sin\tfrac{\pi}{2^6}$, and so on.
There are seven cosines, from $\tfrac{\pi}{2^8}$ up to $\tfrac{\pi}{2^2}$, so the product ends at $\dfrac{1}{2^7}\sin\dfrac{\pi}{2}=\dfrac{1}{128}$.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.