sin^4 (π/(16))+sin^4 ((3π)/(16))+sin^4 ((5π)/(16))+sin^4 ((7π)/(16))


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Question Number 131492 by Ar Brandon last updated on 05/Feb/21

$\sin^{4} \frac{π}{16} + \sin^{4} \frac{3 π}{16} + \sin^{4} \frac{5 π}{16} + \sin^{4} \frac{7 π}{16}$
	Answered by Dwaipayan Shikari last updated on 05/Feb/21

	$s i n \frac{π}{16} = c o s (\frac{π}{2} - \frac{π}{16}) = c o s (\frac{7 π}{16}) s i n (\frac{3 π}{16}) = c o s (\frac{5 π}{16})$ $Φ = {c o s}^{4} \frac{π}{16} + {s i n}^{4} \frac{π}{16} + {s i n}^{4} \frac{3 π}{16} + {c o s}^{4} \frac{3 π}{16}$ $= (1 - \frac{1}{2} {s i n}^{2} \frac{π}{8}) + (1 - \frac{1}{2} {s i n}^{2} \frac{3 π}{8})$ $= 2 - \frac{1}{2} ({s i n}^{2} \frac{π}{8} + {s i n}^{2} \frac{3 π}{8}) = 2 - \frac{1}{2} ({s i n}^{2} \frac{π}{8} + {c o s}^{2} \frac{π}{8}) = 2 - \frac{1}{2} = \frac{3}{2}$
	Commented by Dwaipayan Shikari last updated on 05/Feb/21

	$I t w i l l b e e v e n n i c e r i f y o u t a k e t h e v a l u e o f s i n (\frac{π}{16})^{'} s$ $S e p a r a t e l y : (:)$ $s i n (\frac{π}{16}) = \frac{1}{2} \sqrt{2 + \sqrt{2 - \sqrt{2}}} . . . .$
	Commented by Ar Brandon last updated on 05/Feb/21
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	Answered by Ar Brandon last updated on 05/Feb/21

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