【答】求三角函数式的值：[sin(π/7)]^6-3/8 cos(4π/7)+[cos(π/7)]^6

dodonaomikiki

求解：优美三角函数题目

感觉甚是优美，我把她搞过来啦！

请看图片

Treenewbee

\(sin^6\frac{\pi}{7}+cos^6\frac{\pi}{7}\)

\(=(sin^2\frac{\pi}{7}+cos^2\frac{\pi}{7})(sin^4\frac{\pi}{7}+cos^4\frac{\pi}{7}-sin^2\frac{\pi}{7}cos^2\frac{\pi}{7})\)

\(=1-3sin^2\frac{\pi}{7}cos^2\frac{\pi}{7}=1-\frac{3}{4}sin^2\frac{2\pi}{7}\)

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

故原式值为   \(\frac{5}{8}\)

luyuanhong

楼上 Treenewbee 的解答很好！已收藏。