已知f(α)=sin(π-α)cos(2π-α)sin(-α+3π2)cos(-π-α)cos(-α+3π2)(1)化简f(α);(2)若α是第三象限角,且cos(α-3π2)=15,求f(α)的值.

题目简介

已知f(α)=sin(π-α)cos(2π-α)sin(-α+3π2)cos(-π-α)cos(-α+3π2)(1)化简f(α);(2)若α是第三象限角,且cos(α-3π2)=15,求f(α)的值.

题目详情

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

(1)化简f(α);
(2)若α是第三象限角,且cos(α-
2
)=
1
5
,求f(α)的值.
题型:解答题难度:中档来源:不详

答案

(本题满分12分)
(1)f(α)=
sin(π-α)cos(2π-α)sin(-α+class="stub"3π
2
)
cos(-π-α)cos(-α+class="stub"3π
2
)
=
sinαcosαsin(α-class="stub"π
2
)
cosαcos(α-class="stub"π
2
)
=sinαclass="stub"-cosα
sinα
=-cosα

(2)∵cos(α-class="stub"3π
2
)=cos(α+class="stub"π
2
)=-sinα=class="stub"1
5
,∴sinα=-class="stub"1
5

又α是第三象限角,则cosα=-
1-sin2α
=-
2
6
5
,∴f(α)=
2
6
5

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