问题 解答题
(1)已知tanα=2,求
sin(π-α)cos(2π-α)sin(-α+
2
)
tan(-α-π)sin(-π-α)
的值
(2)已知cos(75°+α)=
1
3
,其中-180°<α<-90°,求sin(105°-α)+cos(375°-α)的值.
答案

(1)原式=

sinαcosα(-cosα)
(-tanα)sinα
(2分)

=

cos2α
tanα
(3分)

tanα=2,

1
cos2α
=1+tan2α=5,

cos2α=

1
5
(6分),∴原式=
1
10
(7分)

(2)原式=sin(75°+α)+cos(15°-α)=2sin(75°+α)(9分)

cos(75°+α)=

1
3
,且-105°<75°+α<-15°,

∴sin(75°+α)<0∴sin(75°+α)=-

1-sin(75°+α)
=-
2
2
3
(12分)

故原式=-

4
3
2
(14分)

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