Question

Sin^4 п/16 + sin^4 3п/16 + sin^4 5п/16 + cos^4 7п/16 вычислите !

Answer 1

$sin^4 \frac{\pi }{16}+sin^4\frac{3\pi }{16}+sin^4\frac{5\pi }{16} +cos^4 \frac{7\pi }{16} =\; [\; sin^4a=(sin^2a)^2= \frac{1-cos2a}{2} \; ]=\\\\=\Big (\frac{1-cos\frac{\pi}{8}}{2} \Big )^2+\Big ( \frac{1-cos\frac{3\pi}{8}}{2} \Big )^2+\Big ( \frac{1-cos\frac{5\pi }{8}}{y} \Big )^2+\Big ( \frac{1+cos\frac{7\pi}{8}}{2}\Big )^2=$

$=\frac{1}{4}\cdot (1-2cos\frac{\pi}{8}+cos^2\frac{\pi}{8}+1-2cos\frac{3\pi}{8}+cos^2\frac{3\pi}{8}+1-2cos \frac{5\pi}{8}+\\\\+cos^2\frac{5\pi}{8} +1+2cos \frac{7\pi}{8}+cos^2\frac{7\pi}{8})=$

$= \frac{1}{4}\cdot (4-2cos\frac{\pi}{8}+\frac{1+cos\frac{\pi}{4}}{2}-2cos\frac{3\pi}{8}+ \frac{1+cos\frac{3\pi}{4}}{2} -2cos\frac{5\pi}{8}+\\\\+\frac{1+cos\frac{5\pi}{4}}{2}+2cos\frac{7\pi}{8}+\frac{1+cos\frac{7\pi}{4}}{2} )=\\\\=\frac{1}{4}\cdot (4+2(cos\frac{7\pi}{8} -cos\frac{\pi}{8})-2(cos\frac{3\pi}{8}+cos\frac{5\pi}{8})+(\frac{1}{2}+\frac{1}{2}\cdot \frac{\sqrt2}{2})+$

$+(\frac{1}{2}+\frac{1}{2}\cdot \frac{-\sqrt2}{2})+(\frac{1}{2}+\frac{1}{2}\cdot \frac{-\sqrt2}{2})+(\frac{1}{2}+\frac{1}{2}\cdot \frac{\sqrt2}{2}))=$

$=\frac{1}{4}\cdot (4+2+2\cdot (-2)\cdot \underbrace {sin\frac{\pi }{2}}_{1}\cdot sin\frac{3\pi }{8}-2\cdot 2\cdot \underbrace{cos\frac{\pi}{2}}_{0}\cdot cos\frac{\pi}{8})=\\\\=\frac{1}{4}\cdot (6-4sin \frac{3\pi }{8} )=\frac{3}{2}-sin\frac{3\pi}{8}=\frac{3}{2}-sin(\frac{\pi}{2}-\frac{\pi}{8})=\\\\=\frac{3}{2}-cos\frac{\pi}{8}=\frac{3}{2}-\frac{\sqrt{2+\sqrt2}}{2}$