[sin^2a-sin^2b]⼀sinacosa-sinbcosb=?演算过程

这个问题是高一必修四的第三章问题,谢谢啊
2024-11-26 04:19:09
推荐回答(2个)
回答1:

[sin^2a-sin^2b]/sinacosa-sinbcosb
=[sin^2-sin^2b]/[(1/2)sina-(1/2)sinb]
=2(sina+sinb)(sina-sinb)/(sina-sinb)
=2(sina+sinb).

回答2:

=sin(a+b)sin(a-b)/[sin(a+b)cos(a-b)]=tg(a+b)