Question

Найдите значение выражения
cos^4 7π/24+sin^4 11π/24+sin^4 17π/24 + cos^4 13π/24​

Answer 1

$cos^4\dfrac{7\pi}{24}+sin^4\dfrac{11\pi}{24}+sin^4\dfrac{17\pi}{24}+cos^4\dfrac{13\pi}{24}=\\\\\\=\Big(\dfrac{1+cos\frac{7\pi}{12}}{2}\Big)^2+\Big(\dfrac{1-cos\frac{11\pi}{12}}{2}\Big)^2+\Big(\dfrac{1-cos\frac{17\pi}{12}}{2}\Big)+\Big(\dfrac{1+cos\frac{13\pi}{12}}{2}\Big)^2=\\\\\\=\dfrac{1}{4}\cdot \Big(1+2cos\dfrac{7\pi}{12}+cos^2\dfrac{7\pi}{12}+1-2cos\dfrac{11\pi}{12}+cos^2\dfrac{11\pi}{12}+\\\\\\+1-2cos\dfrac{17\pi}{12}+cos^2\dfrac{17\pi}{12}+1+2cos\dfrac{13\pi}{12}+cos^2\dfrac{13\pi }{12}\Big)=$

$=\frac{1}{4}\cdot \Big(4+2\cdot 2sin\dfrac{18\pi}{2\cdot 12}\cdot sin\dfrac{4\pi}{2\cdot 12}+2\cdot 2sin\dfrac{30\pi}{2\cdot 12}\cdot sin\dfrac{4\pi}{2\cdot 12}+\\\\\\+\dfrac{1+cos\frac{7\pi}{6}}{2}+\dfrac{1+cos\frac{11\pi}{6}}{2}+\dfrac{1+cos\frac{17\pi}{6}}{2}+\dfrac{1+cos\frac{13\pi}{6}}{2}\Big)=$

$=1+sin\dfrac{3\pi}{4} \cdot sin\dfrac{\pi}{6}+sin\dfrac{5\pi}{4}\cdot sin\dfrac{\pi}{6}+\dfrac{2}{4}-\dfrac{1}{8}\cdot cos\dfrac{\pi}{6}+\dfrac{1}{8}\cdot cos\dfrac{\pi}{6}+\\\\\\+\dfrac{1}{8}\cdot cos\dfrac{5\pi}{6}+\dfrac{1}{8}\cdot cos\dfrac{\pi}{6}=$

$=\dfrac{3}{2}+\dfrac{\sqrt2}{2}\cdot \dfrac{1}{2}-\dfrac{\sqrt2}{2}\cdot \dfrac{1}{2}-\dfrac{1}{8}\cdot \dfrac{\sqrt3}{2}+\dfrac{1}{8}\cdot \dfrac{\sqrt3}{2}-\dfrac{1}{8}\cdot \dfrac{\sqrt3}{2}+\dfrac{1}{8}\cdot \dfrac{\sqrt3}{2}=\dfrac{3}{2}$

$\star \ \ cos^2\alpha =\dfrac{1+cos2\alpha }{2}\ \ ,\ \ \ \ sin^2\alpha =\dfrac{1-cos2\alpha }{2}\ \ ,\\\\\\\star \ \ cos\alpha -cos\beta =2\cdot sin\dfrac{\alpha +\beta }{2}\cdot sin\dfrac{\beta -\alpha }{2}$