Question

prove that 8 sin 10° sin 50° sin70° =1/16​

Answer 1

Correct Statement is

Prove that

$:\implies \tt \: 8sin10 \degree \: sin50\degree \: sin70\degree = 1$

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We know

$\boxed{ \red{\tt \: 2sinxsiny = cos(x - y) - cos(x + y)}}$

$\boxed{ \pink{ \tt \: 2sinxcosy = sin(x + y) + sin(x - y)}}$

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$\large\underline\purple{\bold{Solution :- }}$

Consider LHS

$:\implies \tt \: 8sin10 \degree \: sin50\degree \: sin70\degree$

$:\implies \tt \: 4(2sin50 \degree \: sin10\degree \:) sin70\degree$

$:\implies \tt \: 4(cos(50\degree \: - 10\degree) - cos(50\degree \: + 10\degree))sin70\degree$

$:\implies \tt \: 4(cos40\degree \: - cos60\degree)sin70\degree \:$

$:\implies \tt \: 4(cos40\degree \: - \dfrac{1}{2} )sin70\degree$

$:\implies \tt \: 4cos40\degree \:sin70\degree \: - 2sin70\degree \:$

$:\implies \tt \: 2(2sin70\degree \:cos40\degree \:) - 2sin70\degree \:$

$:\implies \tt \: 2(sin(70\degree \: + 40\degree \:) + sin(70\degree \: - 40\degree \:)) - 2sin70\degree \:$

$:\implies \tt \: 2(sin110\degree \: + sin30\degree \:) - 2sin70\degree \:$

$:\implies \tt \: 2(sin110\degree \: + \dfrac{1}{2} ) - 2sin70\degree \:$

$:\implies \tt \: 2sin110\degree \: + 1 - 2sin(180\degree \: - 110\degree \:)$

$:\implies \tt \: 2sin110\degree \: + 1 - 2sin110\degree \:$

$:\implies \tt \: 1$

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$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$