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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac{3}{16}$
✓

Answer

$\text{LHS}=\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ$
$\sin20^\circ\sin40^\circ\sin80^\circ\times\frac{\sqrt3}{2}$$\Big[\because\ \sin60^\circ=\frac{\sqrt3}{2}\Big]$
$=\ \frac{\sqrt3}{2}\times\frac{1}{2}(2\sin20^\circ\sin40^\circ)\sin80^\circ$
$=\ \frac{\sqrt3}{4}[\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ)]\sin80^\circ$
$=\ \frac{\sqrt3}{4}[\cos20^\circ-\cos60^\circ]\sin80^\circ$
$=\ \frac{\sqrt3}{4}\Big[\cos20^\circ\sin80^\circ-\frac{1}{2}\sin80^\circ\Big]$
$=\ \frac{\sqrt3}{8}[2\cos20^\circ\sin80^\circ-\sin80^\circ]$
$=\ \frac{\sqrt3}{8}[\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ]$
$=\ \frac{\sqrt3}{8}[\sin100^\circ+\sin60^\circ-\sin80^\circ]$
$=\ \frac{\sqrt3}{8}[\sin80^\circ+\sin60^\circ-\sin80^\circ]$
$=\ \frac{\sqrt3}{8}\times\sin60^\circ=\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}$
$=\ \frac{3}{16}=\text{RHS}$

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