Question

Sa se calculeze valoarea expresiei:

$sin^4\frac{pi}{16} +sin^4\frac{3pi}{16} + sin^4\frac{5pi}{16} + sin^4\frac{7pi}{16}$

Answer 1

$\sin^4\dfrac{\pi}{16}+\sin^4\dfrac{3\pi}{16}+\sin^4\dfrac{5\pi}{16}+\sin^4\dfrac{7\pi}{16} = \\ \\ = \sin^4\dfrac{\pi}{16}+\sin^4\dfrac{3\pi}{16}+\sin^4\Big(\dfrac{3\pi}{16}-\dfrac{\pi}{2}\Big)+\sin^4\Big(\dfrac{\pi}{16}-\dfrac{\pi}{2}\Big) =$

$= \sin^4\dfrac{\pi}{16}+\sin^4\dfrac{3\pi}{16}+\cos^4\dfrac{3\pi}{16}+\cos^4\dfrac{\pi}{16} = \\ \\ =\Big(\sin^2 \dfrac{\pi}{16}+\cos^2\dfrac{\pi}{16}\Big)^2 - 2\sin^2\dfrac{\pi}{16}\cos^{2}\dfrac{\pi}{16}+ \\ \\ +\Big(\sin^2 \dfrac{3\pi}{16}+\cos^2\dfrac{3\pi}{16}\Big)^2 - 2\sin^2\dfrac{3\pi}{16}\cos^{2}\dfrac{3\pi}{16} =\\ \\ = 1^2+1^2-\dfrac{\sin^2 \dfrac{\pi}{8}}{2}-\dfrac{\sin^2\dfrac{3\pi}{8}}{2} = \\ \\ = 2-\dfrac{1}{2}\Big[\sin^2 \dfrac{\pi}{8}+\sin^2\Big(\dfrac{\pi}{2}-\dfrac{\pi}{8}\Big)\Big] =$

$=2-\dfrac{1}{2}\Big(\sin^2 \dfrac{\pi}{8}+\cos^2\dfrac{\pi}{8}\Big) = 2-\dfrac{1}{2}\cdot 1 = \boxed{\dfrac{3}{2}}$