首页 > 已知f(α)=sin(2π-α)cos(π+α)cos(π2+α)cos(11π2-α)cos(π-α)sin(3π-α)sin(-π-α)sin(9π2+α)．（Ⅰ）化简f（α）；（Ⅱ）若α是第三象

已知f(α)=sin(2π-α)cos(π+α)cos(π2+α)cos(11π2-α)cos(π-α)sin(3π-α)sin(-π-α)sin(9π2+α)．（Ⅰ）化简f（α）；（Ⅱ）若α是第三象

题目简介

已知f(α)=sin(2π-α)cos(π+α)cos(π2+α)cos(11π2-α)cos(π-α)sin(3π-α)sin(-π-α)sin(9π2+α)．（Ⅰ）化简f（α）；（Ⅱ）若α是第三象

题目详情

已知f(α)=
sin(2π-α)cos(π+α)cos(
π
2
+α)cos(
11π
2
-α)
cos(π-α)sin(3π-α)sin(-π-α)sin(
9π
2
+α)
．
（Ⅰ）化简f（α）；
（Ⅱ）若α是第三象限角，且cos(α-
3π
2
)=
2
2
3
，求f（α）的值．
题型：解答题难度：中档来源：不详

答案

（Ⅰ）f(α)=
sin(2π-α)cos(π+α)cos(class="stub"π
2
+α)cos(class="stub"11π
2
-α)
cos(π-α)sin(3π-α)sin(-π-α)sin(class="stub"9π
2
+α)

=
sinα(-cosα)(-sinα)(-sinα)
(-cosα)sinαsinαcosα
=tanα
（Ⅱ）cos(α-class="stub"3π
2
)=-sinα=
2
2
3
，
∴sinα=-
2
2
3

∵α是第三象限角
∴cosα=-
1-class="stub"8
9
=-class="stub"1
3

∴f（α）=tanα=class="stub"sinα
cosα
=2
2

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