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

题目简介

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

题目详情

已知角α是第三象限角,且f(α)=
sin(π-α)cos(2π-α)tan(-α-π)
tan(π+α)sin(-π-α)

(1)化简f(α);
(2)若cos(α-
2
)=
1
5
,求f(α)的值;
(3)若cos(α+
π
4
)=
3
5
,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(-π-α)
=-class="stub"sinαcosαtanα
tanαsinα
=-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)=-cosα=
2
6
5

(3)∵α是第三象限角
2kπ+class="stub"5
4
π<α<2kπ+class="stub"7
4
π,k∈z

又∵cos(α+class="stub"π
4
)=class="stub"3
5

sin(α+class="stub"π
4
)=-class="stub"4
5

cosα=cos[(α+class="stub"π
4
)-class="stub"π
4
]
=cos(α+class="stub"π
4
)cosclass="stub"π
4
+sin(α+class="stub"π
4
)sinclass="stub"π
4
=-
2
10

∴f(α)=-cosα=
2
10

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