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Transformation Formulae — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae5 Marks
Question
Prove that:
$\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ=\frac{\sqrt3}{16}$
✓

Answer

$\text{LHS}=\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ$
$\Big(\sin10^\circ\sin50^\circ\sin70^\circ\frac{\sqrt3}{2}\Big)$$\Big[\because\ \sin60^\circ=\frac{\sqrt3}{2}\Big]$
$=\ \sin(90^\circ-80^\circ)\sin(90^\circ-40^\circ)\sin(90^\circ-20^\circ)\frac{\sqrt3}{2}$
$=\ \cos80^\circ\cos40^\circ\cos20^\circ\frac{\sqrt3}{2}$
$=\ \frac{\sqrt3}{2\times2}(2\cos40^\circ\cos20^\circ)\cos80^\circ$$[\because\ 2\cos\text{A}\cos\text{B}=(\cos\text{A+B})+\cos(\text{A}-\text{B})]$
$=\ \frac{\sqrt3}{2\times2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20^\circ)]\cos80^\circ$
$=\ \frac{\sqrt3}{2\times2}[\cos60^\circ+\cos20^\circ]\cos80^\circ$
$=\ \frac{\sqrt3}{2\times2}\Big[\frac{1}{2}+\cos20^\circ\Big]\cos80^\circ$
$=\ \frac{\sqrt3}{4}\Big[\frac{1}{2}\cos80^\circ+\cos20^\circ\cos80^\circ\Big]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+2\cos20^\circ\cos80^\circ]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos(80^\circ+20^\circ)+\cos(80^\circ-20^\circ)]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos100^\circ+\cos60^\circ]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ]$
$=\ \frac{\sqrt3}{8}[\cos60^\circ]=\frac{\sqrt3}{16}=\text { RHS}$

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