sin-4-pi-16-sin-4-3pi-16-sin-4-5pi-16-sin-4-7pi-16-

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}}} \dots .

Commented by Ar Brandon last updated on 05/Feb/21

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Answered by Ar Brandon last updated on 05/Feb/21