分类：教育/科学-学习帮助
问题：(sin10度)^2+(sin50度)^2
(sin10度)^2+(sin50度)^2-2sin10度sin50度c120度=?
回答：省略“度”：

(sin10)^+(sin50)^-2sin10sin50cos120
=(sin10)^+(sin50)^-2sin10sin50(-1/2)
=(sin10)^+(sin50)^+sin10sin50
=(1/2)[(1-cos20)+(1-cos100)-(cos60-cos40)]
=(1/2)[3/2-(cos20+cos100)+cos40]
=(1/2)[3/2-2cos60cos40+cos40]
=3/4