If \sin ^2(10^\circ )\sin (20^\circ )\sin (40^\circ )\sin (50^\circ )\sin (70^\circ ) = \alpha - 1 \over 16\sin (10^\circ ) , then 16 + \alpha ^ - 1 is equal to \underline\hspace2cm .

Trigonometric Ratio and Identites PYQ — JEE Main 2022 · 26 Jun · Shift 1 · Q37 with Solution

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Correct answer: 80

We need to evaluate $\sin^2 10^\circ\,\sin 20^\circ\,\sin 40^\circ\,\sin 50^\circ\,\sin 70^\circ$ and compare it with $\alpha-\frac{1}{16}\sin 10^\circ.$
Use complementary-angle identities: $\sin 50^\circ=\cos 40^\circ,\qquad \sin 70^\circ=\cos 20^\circ.$ So the product becomes $\sin^2 10^\circ\,\sin 20^\circ\cos 20^\circ\,\sin 40^\circ\cos 40^\circ.$
Apply $\sin x\cos x=\frac12\sin 2x.$ Thus, $\sin 20^\circ\cos 20^\circ=\frac12\sin 40^\circ,$ $\sin 40^\circ\cos 40^\circ=\frac12\sin 80^\circ.$ Hence

=\frac14\sin^2 10^\circ\sin 40^\circ\sin 80^\circ.$$ 4. Now use the standard identity $$\sin 20^\circ\sin 40^\circ\sin 80^\circ=\frac{\sqrt{3}}{8}.$$ Replacing $20^\circ$ by $10^\circ$ scaled form is not directly useful, so instead use $$\sin 40^\circ\sin 80^\circ=2\cos 20^\circ\sin 20^\circ\sin 80^\circ?$$ That is messy. A cleaner route is: Use $$\sin 3\theta=4\sin\theta\sin(60^\circ+\theta)\sin(60^\circ-\theta).$$ For $\theta=10^\circ$, $$\sin 30^\circ=4\sin 10^\circ\sin 70^\circ\sin 50^\circ.$$ Since $\sin 30^\circ=\frac12$, $$\sin 10^\circ\sin 50^\circ\sin 70^\circ=\frac18.$$ Therefore $$\sin^2 10^\circ\sin 20^\circ\,\sin 40^\circ\sin 50^\circ\sin 70^\circ =\frac18\sin 10^\circ\sin 20^\circ\sin 40^\circ.$$ 5. Again use the identity $$\sin 20^\circ\sin 40^\circ\sin 80^\circ=\frac{\sqrt3}{8},$$ and since $\sin 80^\circ=2\sin 40^\circ\cos 40^\circ$, this is not simplest. Better use another standard identity: $$\sin 3\theta=4\sin\theta\sin(60^\circ+\theta)\sin(60^\circ-\theta).$$ Now put $\theta=20^\circ$: $$\sin 60^\circ=4\sin 20^\circ\sin 80^\circ\sin 40^\circ.$$ So $$\frac{\sqrt3}{2}=4\sin 20^\circ\sin 40^\circ\sin 80^\circ.$$ Hence $$\sin 20^\circ\sin 40^\circ\sin 80^\circ=\frac{\sqrt3}{8}.$$ Also $\sin 80^\circ=\cos 10^\circ$. Now use $$2\sin 10^\circ\cos 10^\circ=\sin 20^\circ$$ so $$\cos 10^\circ=\frac{\sin 20^\circ}{2\sin 10^\circ}.$$ Then $$\sin 20^\circ\sin 40^\circ\cos 10^\circ=\frac{\sqrt3}{8}$$ becomes $$\sin 20^\circ\sin 40^\circ\cdot \frac{\sin 20^\circ}{2\sin 10^\circ}=\frac{\sqrt3}{8},$$ which is still not ideal. So let us use the direct well-known identity: $$\sin 20^\circ\sin 40^\circ=\sin 10^\circ\sin 50^\circ.$$ And since $$\sin 50^\circ=\cos 40^\circ,$$ this gives a simplification route; but the cleanest standard result is $$\sin 10^\circ\sin 20^\circ\sin 40^\circ=\frac{\sqrt3}{8}\sin 10^\circ\sec 10^\circ?$$ Instead, compute directly: Using product-to-sum, $$\sin 20^\circ\sin 40^\circ=\frac12\big(\cos 20^\circ-\cos 60^\circ\big) =\frac12\left(\cos 20^\circ-\frac12\right).$$ Thus $$P=\frac18\sin 10^\circ\cdot \frac12\left(\cos 20^\circ-\frac12\right) =\frac{1}{16}\sin 10^\circ\cos 20^\circ-\frac{1}{32}\sin 10^\circ.$$ Now $$\cos 20^\circ=1-2\sin^2 10^\circ.$$ So $$P=\frac{1}{16}\sin 10^\circ(1-2\sin^2 10^\circ)-\frac{1}{32}\sin 10^\circ =\frac{1}{32}\sin 10^\circ-\frac18\sin^3 10^\circ.$$ This form is not matching the target immediately, so let us use triple-angle: $$\sin 30^\circ=3\sin 10^\circ-4\sin^3 10^\circ=\frac12.$$ Hence $$4\sin^3 10^\circ=3\sin 10^\circ-\frac12,$$ $$\sin^3 10^\circ=\frac34\sin 10^\circ-\frac18.$$ Substitute into $P$: $$P=\frac{1}{32}\sin 10^\circ-\frac18\left(\frac34\sin 10^\circ-\frac18\right) =\frac{1}{32}\sin 10^\circ-\frac{3}{32}\sin 10^\circ+\frac{1}{64}.$$ Therefore $$P=\frac{1}{64}-\frac{1}{16}\sin 10^\circ.$$ 6. Comparing with $$P=\alpha-\frac{1}{16}\sin 10^\circ,$$ we get $$\alpha=\frac{1}{64}.$$ So $$\alpha^{-1}=64.$$ Therefore $$16+\alpha^{-1}=16+64=80.$$ 7. Final answer: $$\boxed{80}$$ The derived answer matches the stored correct answer.

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View written solutionFree

Correct answer: 80

We need to evaluate $\sin^2 10^\circ\,\sin 20^\circ\,\sin 40^\circ\,\sin 50^\circ\,\sin 70^\circ$ and compare it with $\alpha-\frac{1}{16}\sin 10^\circ.$
Use complementary-angle identities: $\sin 50^\circ=\cos 40^\circ,\qquad \sin 70^\circ=\cos 20^\circ.$ So the product becomes $\sin^2 10^\circ\,\sin 20^\circ\cos 20^\circ\,\sin 40^\circ\cos 40^\circ.$
Apply $\sin x\cos x=\frac12\sin 2x.$ Thus, $\sin 20^\circ\cos 20^\circ=\frac12\sin 40^\circ,$ $\sin 40^\circ\cos 40^\circ=\frac12\sin 80^\circ.$ Hence

=\frac14\sin^2 10^\circ\sin 40^\circ\sin 80^\circ.$$ 4. Now use the standard identity $$\sin 20^\circ\sin 40^\circ\sin 80^\circ=\frac{\sqrt{3}}{8}.$$ Replacing $20^\circ$ by $10^\circ$ scaled form is not directly useful, so instead use $$\sin 40^\circ\sin 80^\circ=2\cos 20^\circ\sin 20^\circ\sin 80^\circ?$$ That is messy. A cleaner route is: Use $$\sin 3\theta=4\sin\theta\sin(60^\circ+\theta)\sin(60^\circ-\theta).$$ For $\theta=10^\circ$, $$\sin 30^\circ=4\sin 10^\circ\sin 70^\circ\sin 50^\circ.$$ Since $\sin 30^\circ=\frac12$, $$\sin 10^\circ\sin 50^\circ\sin 70^\circ=\frac18.$$ Therefore $$\sin^2 10^\circ\sin 20^\circ\,\sin 40^\circ\sin 50^\circ\sin 70^\circ =\frac18\sin 10^\circ\sin 20^\circ\sin 40^\circ.$$ 5. Again use the identity $$\sin 20^\circ\sin 40^\circ\sin 80^\circ=\frac{\sqrt3}{8},$$ and since $\sin 80^\circ=2\sin 40^\circ\cos 40^\circ$, this is not simplest. Better use another standard identity: $$\sin 3\theta=4\sin\theta\sin(60^\circ+\theta)\sin(60^\circ-\theta).$$ Now put $\theta=20^\circ$: $$\sin 60^\circ=4\sin 20^\circ\sin 80^\circ\sin 40^\circ.$$ So $$\frac{\sqrt3}{2}=4\sin 20^\circ\sin 40^\circ\sin 80^\circ.$$ Hence $$\sin 20^\circ\sin 40^\circ\sin 80^\circ=\frac{\sqrt3}{8}.$$ Also $\sin 80^\circ=\cos 10^\circ$. Now use $$2\sin 10^\circ\cos 10^\circ=\sin 20^\circ$$ so $$\cos 10^\circ=\frac{\sin 20^\circ}{2\sin 10^\circ}.$$ Then $$\sin 20^\circ\sin 40^\circ\cos 10^\circ=\frac{\sqrt3}{8}$$ becomes $$\sin 20^\circ\sin 40^\circ\cdot \frac{\sin 20^\circ}{2\sin 10^\circ}=\frac{\sqrt3}{8},$$ which is still not ideal. So let us use the direct well-known identity: $$\sin 20^\circ\sin 40^\circ=\sin 10^\circ\sin 50^\circ.$$ And since $$\sin 50^\circ=\cos 40^\circ,$$ this gives a simplification route; but the cleanest standard result is $$\sin 10^\circ\sin 20^\circ\sin 40^\circ=\frac{\sqrt3}{8}\sin 10^\circ\sec 10^\circ?$$ Instead, compute directly: Using product-to-sum, $$\sin 20^\circ\sin 40^\circ=\frac12\big(\cos 20^\circ-\cos 60^\circ\big) =\frac12\left(\cos 20^\circ-\frac12\right).$$ Thus $$P=\frac18\sin 10^\circ\cdot \frac12\left(\cos 20^\circ-\frac12\right) =\frac{1}{16}\sin 10^\circ\cos 20^\circ-\frac{1}{32}\sin 10^\circ.$$ Now $$\cos 20^\circ=1-2\sin^2 10^\circ.$$ So $$P=\frac{1}{16}\sin 10^\circ(1-2\sin^2 10^\circ)-\frac{1}{32}\sin 10^\circ =\frac{1}{32}\sin 10^\circ-\frac18\sin^3 10^\circ.$$ This form is not matching the target immediately, so let us use triple-angle: $$\sin 30^\circ=3\sin 10^\circ-4\sin^3 10^\circ=\frac12.$$ Hence $$4\sin^3 10^\circ=3\sin 10^\circ-\frac12,$$ $$\sin^3 10^\circ=\frac34\sin 10^\circ-\frac18.$$ Substitute into $P$: $$P=\frac{1}{32}\sin 10^\circ-\frac18\left(\frac34\sin 10^\circ-\frac18\right) =\frac{1}{32}\sin 10^\circ-\frac{3}{32}\sin 10^\circ+\frac{1}{64}.$$ Therefore $$P=\frac{1}{64}-\frac{1}{16}\sin 10^\circ.$$ 6. Comparing with $$P=\alpha-\frac{1}{16}\sin 10^\circ,$$ we get $$\alpha=\frac{1}{64}.$$ So $$\alpha^{-1}=64.$$ Therefore $$16+\alpha^{-1}=16+64=80.$$ 7. Final answer: $$\boxed{80}$$ The derived answer matches the stored correct answer.

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