问题
解答题
(1)化简:
(2)求证:
|
答案
(1)
sin(2π-α)sin(π+α)cos(-π-α) |
sin(3π-α)cos(π-α) |
-sinα•(-sinα)•(-cosα) |
sinα•(-cosα) |
(2)证明:∵
cosx |
1-sinx |
1+sinx |
cosx |
cos2x-(1+sinx)(1-sinx) |
(1-sinx)•cosx |
=
cos2x-(1-sin2x) |
(1-sinx)•cosx |
=0.
∴
cosx |
1-sinx |
1+sinx |
cosx |