问题
解答题
已知sin
(Ⅰ)求tanx的值; (Ⅱ)求
|
答案
(1)由sin
-2cosx 2
=0,⇒tanx 2
=2,x 2
∴tanx=
=2tan x 2 1-tan2 x 2
=-2×2 1-22
.4 3
(2)原式=
=cos2x-sin2x
(2
cosx-2 2
sinx)sinx2 2
由(1)知cosx-sinx≠0(cosx-sinx)(cosx+sinx) (cosx-sinx)sinx
所以上式=
=cotx+1=(-cosx+sinx sinx
)+1=3 4
.1 4