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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove that $\sin 20^{\large\circ}\sin 40^{\large\circ}\sin 80^{\large\circ}=\large\frac{\sqrt 3}{8}$

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Toolbox:
  • $2\sin A\sin B=\cos(A-B)-\cos(A+B)$
  • $2\cos A\sin B=\sin(A+B)-\sin(A-B)$
LHS
$\sin 20^{\large\circ}\sin 40^{\large\circ}\sin 80^{\large\circ}$
$\Rightarrow \large\frac{1}{2}$$(\cos 40^{\large\circ}-\cos 120^{\large\circ})\sin 20^{\large\circ}$
$\Rightarrow \large \frac{1}{4}$$(2\cos 40^{\large\circ}\sin 20^{\large\circ}-2\cos 120^{\large\circ}\sin 120^{\large\circ}$
$\Rightarrow \large\frac{1}{4}$$[\sin (40+20)^{\large\circ}-\sin (40-20)^{\large\circ}-2(-\large\frac{1}{2})$$\sin 20^{\large\circ}]$
$\Rightarrow \large\frac{1}{4}$$[\sin 60^{\large\circ}-\sin 20^{\large\circ}+\sin 20^{\large\circ}]$
$\Rightarrow \large\frac{1}{4}$$\sin 60^{\large\circ}$
$\Rightarrow \large\frac{1}{4}.\frac{\sqrt 3}{2}=\frac{\sqrt 3}{8}$
Hence proved
answered May 21, 2014 by sreemathi.v
 

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