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← JEE Main 2022 · 26 Jun · Shift 1
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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 26

If sin⁡2(10°)sin⁡(20°)sin⁡(40°)sin⁡(50°)sin⁡(70°)=α−116sin⁡(10°)\sin^2(10°)\sin(20°)\sin(40°)\sin(50°)\sin(70°)=\alpha-\dfrac1{16}\sin(10°), then 16+α−116+\alpha^{-1} is equal to ______.

Answer and worked solution

Official answer: 80

Pair up. sin⁡20cos⁡20=12sin⁡40\sin20\cos20=\frac12\sin40 and sin⁡40cos⁡40=12sin⁡80\sin40\cos40=\frac12\sin80, so the product is 14sin⁡210°sin⁡40°sin⁡80°\frac14\sin^210°\sin40°\sin80°.

Simplify. sin⁡40sin⁡80=12(cos⁡40+12)\sin40\sin80=\frac12\left(\cos40+\frac12\right); using sin⁡10cos⁡40=12(sin⁡50−12)\sin10\cos40=\frac12(\sin50-\frac12), sin⁡50+sin⁡10=cos⁡20\sin50+\sin10=\cos20 and sin⁡10cos⁡20=14−12sin⁡10\sin10\cos20=\frac14-\frac12\sin10, the product becomes 164−116sin⁡10°\dfrac1{64}-\dfrac1{16}\sin10°.

So α=164\alpha=\frac1{64} and 16+α−1=8016+\alpha^{-1}=80.

Answer: 80

Topics:Trigonometric ratios and identities
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