1.计算cos 24°cos 36°-cos 66°cos 54°的结果为( )
A.0 B.1
2
C.32 D.-12
解析:选B.原式=cos 24°cos 36°-sin 24°sin 36°
=cos(24°+36°)=cos 60°=1
2
.
2.若α∈(π4ππ
2,π),且sin α=5,则sin(α+4)+cos(α+4
)=(A.425 B.-425
C.325 D.-325
解析:选D.∵sin α=4π
5,2
<α<π,
∴cos α=-3
5.
∴sin(α+π4)+cos(α+π4)=2sin(α+π
2
)
=2cos α=-32
5
.
3.(2013·山西考前适应性训练)sin 20°cos 20°
cos 50°
=( )
A.2 B.22 C.2 D.1
2
11
解析:选D.sin 20°cos 20°2sin 40°2sin 40°cos 50°=cos 50°=1
sin 40°=2
.
4.(2013·沈阳四校联考)若1+cos 2α1
sin 2α=2
,则tan 2α等于( A.554 B.-4 C.443 D.-3 解析:选D.1+cos 2αsin 2α=2cos2α2sin αcos α=cos α1sin α=2
,
∴tan α=2,∴tan 2α=2tan α44
1-tan2α=1-4=-3,故选D. tan?π
+α?·cos 2α
5.(2013·石家庄质检)计算4
的值为( )
2cos2?π4
-α?A.-2 B.2 C.-1 D.1
)
)
人教A版数学必修四3.2简单的三角恒等变换基础达标(含答案解析)
1.计算cos24°cos36°-cos66°cos54°的结果为()A.0B.12C.32D.-12解析:选B.原式=cos24°cos36°-sin24°sin36°=cos(24°+36°)=cos60°=12
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