诱导公式练习题(2)
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基 础 巩 固
一、选择题
1.若cos65°=a ,则sin25°的值是( ) A .-a B .a C.1-a 2 D .-1-a 2
[答案] B
2.若sin(π2+θ)<0,且cos(π
2-θ)>0,则θ是( ) A .第一象限角 B .第二象限角 C .第三象限角 D .第四象限角
[答案] B
3.已知cos ⎝
⎛⎭
⎪⎫π2+α=-3
5,且α是第二象限角,则sin ⎝
⎛⎭
⎪⎫α-3π2的结
果是( )
A.45 B .-4
5 C .±45 D.35
[答案] B
[解析] ∵cos ⎝ ⎛⎭
⎪⎫π2+α=-3
5, ∴-sin α=-35,∴sin α=3
5, 又α是第二象限角,∴cos α=-4
5,
∴sin ⎝ ⎛⎭
⎪⎫α-3π2=cos α=-4
5.
4.已知sin α=35,则sin(π
2+α)的值为( ) A .-3
5 B .-4
5 C.4
5 D .±45
[答案] D
[解析] sin(π2+α)=cos α,而sin α=3
5, ∴cos α=±45,于是sin(π2+α)=±4
5.
5.已知sin(α+π4)=13,则cos(π
4-α)的值为( ) A.22
3 B .-22
3 C.1
3 D .-13 [答案] C
[解析] cos(π4-α)=cos[π2-(π
4+α)]. =sin(α+π4)=1
3.
6.已知cos(3π2+α)=-3
5,且α是第四象限角,则cos(-3π+α)( )
A.45 B .-4
5 C .±45
D.35
[答案] B
[解析] ∵cos(3π2+α)=-35,∴sin α=-3
5, ∴cos(-3π+α)=-cos α=-1-sin 2
α=-4
5.
二、填空题
7.化简sin (15π2+α)cos (α-π2)
sin (9π2-α)cos (3π
2+α)=________. [答案] -1 [解析] 原式
=sin[8π+(α-π2)]cos (π
2-α)sin[4π+(π2-α)]cos[π+(π
2+α)]
=
sin (α-π
2)sin α
sin (π2-α)[-cos (π
2+α)]
=-cos αsin α
cos α[-(-sin α)]
=-1.
8.已知sin(α-π4)=35,那么cos(α+π
4)的值是__________. [答案] -3
5
[解析] ∵(α+π4)-(α-π4)=π
2, ∴α+π4=π2+(α-π4),
∴cos(α+π4)=cos[π2+(α-π4)]=-sin(α-π4)=-3
5. 三、解答题
9.化简:sin (2π+α)cos (π-α)cos (π2-α)cos (7π
2-α)
cos (π-α)sin (3π-α)sin (-π+α)sin (5π
2+α). [解析] 原式=
sin α(-cos α)sin αcos[2π+(π+π
2-α)]
-cos αsin[2π+(π-α)]sin[-(π-α)]sin[2π+(π
2+α)] =sin αsin αcos[π+(π
2-α)]sin (π-α)[-sin (π-α)]sin (π2+α)
=sin αsin α[-cos (π
2-α)]sin α(-sin α)cos α=sin α(-sin α)(-sin α)cos α=tan α.
10.已知角α的终边经过点P (-4,3), 求
cos (π
2+α)sin (-π-α)
cos (11π2-α)sin (9π
2+α)
的值. [解析] ∵角α的终边经过点P (-4,3), ∴tan α=y x =-3
4.
∴原式=-sin α·sin α
-sin α·cos α
=tan α=-3
4.
能 力 提 升
一、选择题
1.已知sin(5π2+α)=1
5,那么cos α=( ) A .-25 B .-15 C.15 D.25
[答案] C
[解析] 本题考查诱导公式,由sin(π2+α)=cos α=1
5,知选C. 2.已知sin α=513,则cos(π
2+α)等于( ) A.513 B.1213 C .-513 D .-1213 [答案] C
[解析] cos(π2+α)=-sin α=-5
13.
3.若sin(3π+α)=-12,则cos(7π
2-α)等于( ) A .-12 B.12 C.32 D .-3
2 [答案] A
[解析] 由已知,得sin α=1
2, 则cos(7π2-α)=-sin α=-12.
4.若sin(π3-α)=13,则cos(5π
6-α)的值为( ) A.13 B .-13 C.223 D .-223
[答案] B
[解析] cos(5π6-α)=cos[π2+(π
3-α)] =-sin(π3-α)=-1
3.
5.已知sin(α+π2)=13,α∈(-π
2,0),则tan α等于( ) A .-2 2 B .2 2 C .-2
4 D.24 [答案] A
[解析] sin(α+π2)=cos α=13,又α∈(-π
2,0), 所以sin α=-
1-cos 2α=-22
3,
则tan α=sin α
cos α=-2 2.
6.若sin α+cos αsin α-cos α=2,sin(α-5π)·sin(3π2-α)等于( )
A.34
B.3
10 C .±310 D .-310
[答案] B