(1)化简:sin(2π-α)sin(π+α)cos(-π-α)sin(3π-α)cos(π-α)(2)求证:cosx1-sinx=1+sinxcosx.-数学

题目简介

(1)化简:sin(2π-α)sin(π+α)cos(-π-α)sin(3π-α)cos(π-α)(2)求证:cosx1-sinx=1+sinxcosx.-数学

题目详情

(1)化简:
sin(2π-α)sin(π+α)cos(-π-α)
sin(3π-α)cos(π-α)

(2)求证:
cosx
1-sinx
=
1+sinx
cosx
题型:解答题难度:中档来源:不详

答案

(1)
sin(2π-α)sin(π+α)cos(-π-α)
sin(3π-α)cos(π-α)
=
-sinα•(-sinα)•(-cosα)
sinα•(-cosα)
=sinα;
  (2)证明:∵class="stub"cosx
1-sinx
-class="stub"1+sinx
cosx

cos2x-(1+sinx)(1-sinx)
(1-sinx)•cosx

=
cos2x-(1-sin2x)
(1-sinx)•cosx

=0.
class="stub"cosx
1-sinx
=class="stub"1+sinx
cosx

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