If sin²10°sin20°sin40°sin50°sin70°=α−1/16sin10°, Find 16+1/α

JEE Main 2022 — 26 June, Shift 1. Previous Year Question.

Problem

If $\sin^2 10^\circ \sin 20^\circ \sin 40^\circ \sin 50^\circ \sin 70^\circ = \alpha – \dfrac{1}{16}\sin 10^\circ$, then $16+\alpha^{-1}$ is equal to:

Key insight. Five different sine terms look unmanageable at first, but converting $\sin 70^\circ$ and $\sin 50^\circ$ into $\cos 20^\circ$ and $\cos 40^\circ$ (complementary angles) pairs each cosine with the matching sine right next to it — and each such pair collapses using $\sin\theta\cos\theta = \tfrac{1}{2}\sin 2\theta$, repeatedly, until only $\sin 10^\circ$ terms remain.

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Approach

The expression mixes sines of $10^\circ, 20^\circ, 40^\circ, 50^\circ, 70^\circ$ — an unpromising-looking set until noticing that $70^\circ = 90^\circ-20^\circ$ and $50^\circ=90^\circ-40^\circ$, so $\sin 70^\circ=\cos20^\circ$ and $\sin50^\circ=\cos40^\circ$. Pairing each of these cosines with its neighbouring sine and repeatedly applying $\sin\theta\cos\theta=\tfrac12\sin2\theta$ steadily halves the angles involved, funnelling everything toward a single $\sin 10^\circ$ term.

Solution

Step 1 — Convert to complementary angles

$$\sin 70^\circ = \cos 20^\circ, \qquad \sin 50^\circ = \cos 40^\circ$$

So the product becomes:

$$\sin^2 10^\circ \cdot (\sin 20^\circ \cos 20^\circ) \cdot (\sin 40^\circ \cos 40^\circ)$$

Step 2 — Collapse each sin–cos pair

Using $\sin\theta\cos\theta = \tfrac12\sin 2\theta$ on both pairs:

$$\sin 20^\circ\cos 20^\circ = \frac{\sin 40^\circ}{2}, \qquad \sin 40^\circ\cos 40^\circ = \frac{\sin 80^\circ}{2}$$

Substituting back:

$$\sin^2 10^\circ \cdot \frac{\sin 40^\circ}{2}\cdot\sin 40^\circ\cdot\cos 40^\circ = \sin^2 10^\circ\cdot\frac{\sin 40^\circ}{2}\cdot\frac{\sin 80^\circ}{2} = \frac{1}{4}\sin^2 10^\circ\sin 40^\circ \sin 80^\circ$$

Step 3 — Convert $\sin 80^\circ$ and collapse again

$\sin 80^\circ = \cos 10^\circ$, so pairing it with the remaining $\sin 10^\circ$:

$$\frac{1}{4}\sin 10^\circ\cdot(\sin 10^\circ\cos 10^\circ)\cdot\sin 40^\circ = \frac{1}{4}\sin 10^\circ\cdot\frac{\sin 20^\circ}{2}\cdot\sin 40^\circ = \frac{1}{8}\sin 10^\circ\sin 20^\circ\sin 40^\circ$$

Step 4 — One more collapse using $\sin 20^\circ = 2\sin10^\circ\cos10^\circ$… instead, pair $\sin20^\circ$ and $\sin40^\circ$ via product-to-sum

Using $2\sin A\sin B = \cos(A-B)-\cos(A+B)$ on $\sin 20^\circ$ and $\sin 40^\circ$:

$$2\sin 20^\circ \sin 40^\circ = \cos 20^\circ – \cos 60^\circ = \cos 20^\circ – \frac{1}{2}$$

So $\sin 20^\circ\sin 40^\circ = \dfrac{1}{2}\cos 20^\circ – \dfrac{1}{4}$, and the expression becomes:

$$\frac{1}{8}\sin 10^\circ\left(\frac{1}{2}\cos 20^\circ – \frac{1}{4}\right) = \frac{1}{16}\sin 10^\circ\cos 20^\circ – \frac{1}{32}\sin 10^\circ$$

Step 5 — Expand $\sin 10^\circ \cos 20^\circ$ using product-to-sum

Using $2\sin A\cos B = \sin(A+B) – \sin(B-A)$:

$$2\sin 10^\circ\cos 20^\circ = \sin 30^\circ – \sin 10^\circ = \frac{1}{2} – \sin 10^\circ$$

So $\sin 10^\circ\cos 20^\circ = \dfrac{1}{4} – \dfrac{1}{2}\sin 10^\circ$. Substituting back:

$$\frac{1}{16}\left(\frac{1}{4}-\frac{1}{2}\sin 10^\circ\right) – \frac{1}{32}\sin 10^\circ = \frac{1}{64} – \frac{1}{32}\sin 10^\circ – \frac{1}{32}\sin 10^\circ = \frac{1}{64} – \frac{1}{16}\sin 10^\circ$$

Step 6 — Identify $\alpha$

Comparing $\dfrac{1}{64} – \dfrac{1}{16}\sin 10^\circ$ with $\alpha – \dfrac{1}{16}\sin 10^\circ$:

$$\alpha = \frac{1}{64}$$

Step 7 — Compute $16 + \alpha^{-1}$

$$\alpha^{-1} = 64, \qquad 16 + 64 = 80$$

Answer

Common mistakes

• Losing track of which complementary pair to use at each stage — with five different angles in play, it’s easy to convert the wrong term or apply an identity to the wrong pair, especially since several product-to-sum steps are needed in sequence.
• Forgetting that the answer needs $\alpha^{-1}$, not $\alpha$ itself — after all that simplification, mixing up $\alpha=\tfrac{1}{64}$ with $\alpha^{-1}=64$ at the last step throws away all the preceding work.

Practise next

• Evaluate $\sin10^\circ\sin30^\circ\sin50^\circ\sin70^\circ$, using the same complementary-angle and product-to-sum approach.