已知f(α)=sin(α-3π)cos2(2π-α)sin(-α+32π)cos(-π-α)sin(-π-α)sin(π2+α).(1)化简f(α);(2)若α=-91π3,求f(α)的值.-数学

题目简介

已知f(α)=sin(α-3π)cos2(2π-α)sin(-α+32π)cos(-π-α)sin(-π-α)sin(π2+α).(1)化简f(α);(2)若α=-91π3,求f(α)的值.-数学

题目详情

已知f(α)=
sin(α-3π)cos2(2π-α)sin(-α+
3
2
π)
cos(-π-α)sin(-π-α)sin(
π
2
+α)

(1)化简f(α);
(2)若α=-
91π
3
,求f(α)的值.
题型:解答题难度:中档来源:不详

答案

解(1)∵f(α)=
sin(α-3π)cos2(2π-α)sin(-α+class="stub"3
2
π)
cos(-π-α)sin(-π-α)sin(class="stub"π
2
+α)

=
-sinαcos2α•(-cosα)
-cosα•sinα•cosα

=-cosα.
∴f(α)=-cosα.
(2)∵f(α)=-cosα
∴当α=-class="stub"91π
3
时,f(α)=-cos(-class="stub"91π
3

=-cos(30π+class="stub"π
3

=-cosclass="stub"π
3

=class="stub"1
2

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