化简三角函数式
化简(sinα)^2+(sinβ)^2-(sinα)^2*(sinβ)^2+(cosα)^2*(cosβ)^2 要过程,谢谢。
(sinα)^2+(sinβ)^2-(sinα)^2*(sinβ)^2+(cosα)^2*(cosβ)^2 =(sinα)^2+(sinβ)^2-(sinα)^2*(sinβ)^2+ [1-(sinα)^2+]*[1-(sinβ)^2] =(sinα)^2+(sinβ)^2-(sinα)^2*(sinβ)^2 +1-[(sinα)^2+(sinβ)^2]+(sinα)^2*(sinβ)^ =1
答:楼上答案都有点问题,直到最后一步之前的过程是对的,错在不该分类讨论和有两个答案,实质上α已经由sinα+cosα=(√3+1)/2 ① , sinαcosα ...详情>>
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