化简:tan(3π-α)sin(π+α)sin(3π2-α)+sin(2π-α)cos(α-7π2)sin(3π2+α)cos(2π+α).-数学

题目简介

化简:tan(3π-α)sin(π+α)sin(3π2-α)+sin(2π-α)cos(α-7π2)sin(3π2+α)cos(2π+α).-数学

题目详情

化简:
tan(3π-α)
sin(π+α)sin(
2
-α)
+
sin(2π-α)cos(α-
2
)
sin(
2
+α)cos(2π+α)
题型:解答题难度:中档来源:不详

答案

∵tan(3π-α)=-tan α,sin(π-α)=sin α,sin(2π-α)=-sin α,cos(2π+α)=cos α,sin(class="stub"3π
2
-α)=-cos α,
cos(α-class="stub"7π
2
)=cos(class="stub"7π
2
-α)=cos(4π-class="stub"π
2
-α)=cos(class="stub"π
2
+α)=-sin α,sin(class="stub"3π
2
+α)=-cos α,
∴原式=class="stub"-tanα
sinα•(-cosα)
+
-sinα•(-sinα)
-cosα•cosα

=class="stub"1
cos2α
-
sin2α
cos2α

=
1-sin2α
cos2α

=
cos2α
cos2α
=1.

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