若sin(α-π)=2cos(2π-α),求sin(2π-α)cos(α+π)cos(α+π2)cos(11π2-α)cos(π-α)sin(3π-α)sin(-α-π)sin(9π2+α).-数学

题目简介

若sin(α-π)=2cos(2π-α),求sin(2π-α)cos(α+π)cos(α+π2)cos(11π2-α)cos(π-α)sin(3π-α)sin(-α-π)sin(9π2+α).-数学

题目详情

若sin(α-π)=2cos(2π-α),求
sin(2π-α)cos(α+π)cos(α+
π
2
)cos(
11π
2
-α)
cos(π-α)sin(3π-α)sin(-α-π)sin(
2
+α)
题型:解答题难度:中档来源:不详

答案

∵sin(α-π)=2cos(2π-α),
∴-sinα=2cosα,即tanα=-2,
则原式=
(-sinα)(-cosα)(-sinα)(-sinα)
(-cosα)sinαsinαcosα
=-tanα=2.

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