sin(π-α)cos(π2+α)sin(π+α)+sin(π2-α)cos(π2-α)cos(π+α)=______.-数学

题目简介

sin(π-α)cos(π2+α)sin(π+α)+sin(π2-α)cos(π2-α)cos(π+α)=______.-数学

题目详情

sin(π-α)cos(
π
2
+α)
sin(π+α)
+
sin(
π
2
-α)cos(
π
2
-α)
cos(π+α)
=______.
题型:填空题难度:偏易来源:不详

答案

原式=
sinα•(-sinα)
-sinα
+class="stub"cosαsinα
-cosα

=
-sin2α
-sinα
+class="stub"cosαsinα
-cosα

=sinα-sinα=0.
故答案为:0

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