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

题目简介

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

题目详情

已知f(α)=sin(π-α)cos(2π-α)tan(-α-π)tan(-π+α)sin(-α-π)
(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
(3)若α=-1860°,求f(α)的值.
题型:解答题难度:中档来源:不详

答案

(1)∵f(α)=sin(π-α)cos(2π-α)tan(-α-π)tan(-π+α)sin(-α-π)
=sinαcosα•(-tanα)[-(-tanα)]•sinα
=-
sin4α
cosα

(2)∵cos(α-class="stub"3π
2
)=class="stub"1
5

∴sinα=-class="stub"1
5

又α是第三象限角,
∴cosα=-
2
6
5

∴f(α)=
6
1500

(3)∵α=-1860°,
∴f(α)=f(-1860°)=-
sin4(-1860°)
cos(-1860°)
=
(-
3
2
)
4
class="stub"1
2
=-class="stub"9
8

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