课时作业(三十一) 诱导公式(二)
[练基础]
π1
+A?等于( ) 1.如果cos(π+A)=-,那么sin??2?2
11A.- B.
2233
C.- D. 22
π
θ-?相等的是( ) 2.下列式子与sin??2?
π?π
+θ B.cos?+θ? A.sin??2??2?33
π-θ? D.sin?π+θ? C.cos??2??2?
π?sin??2+θ?-cos?π-θ?
3.已知tan θ=2,则等于( )
π??sin?2+θ?-sin?π-θ?
A.2 B.-2
2
C.0 D.
3
5π5
α+?=________. 4.若cos α=-,且α是第三象限角,则cos?2??13
3π
-α?cos?6π-α?tan?2π-α?cos??2?
5.求=________.
3π??3π??sin?α+2?cos?α+2?4cos α-sin α1
6.已知=.
3sin α+2cos α4
(1)求tan α的值;
3π?
(2)求sin(π-α)sin??2-α?的值.
[提能力]
7.(多选)若角A,B,C是△ABC的三个内角,则下列等式中不成立的是( ) A.cos(A+B)=cos C B.sin(A+B)=-sin C
A+CC.cos=sin B
2B+CAD.sin=cos 22
π?1
8.若sin θ+cos θ=,且θ∈(0,π),则sin(π+θ)+sin??2+θ?=________. 59.化简:
cos?α-π??π?π+α?; (1)·sin?α-2?cos??2?sin?π-α?
π3π
α-?-sin?+α?cos(α-2π). (2)sin(-α-5π)cos??2??2?
[战疑难]
π15ππ
x+?=,则sin?-x?+sin2?-x?=________. 10.已知sin??6?4?6??3?
课时作业(三十一) 诱导公式(二) 11.解析:cos(π+A)=-cos A=-2, 1∴cos A=2, ?π?1??+A∴sin2=cos A=2. ??答案:B π???π?2.解析:因为sin?θ-2?=-sin?2-θ?=-cos θ, ?????π?对于A,sin?2+θ?=cos θ; ???π??对于B,cos2+θ?=-sin θ; ???3π???π??对于C,cos?2-θ?=cos?π+?2-θ?? ???????π??=-cos2-θ?=-sin θ; ???3π???π??对于D,sin?2+θ?=sin?π+?2+θ?? ???????π??=-sin2+θ?=-cos θ. ??答案:D ?π??sin2+θ?-cos?π-θ?cos θ+cos θ22??3.解析:====?π?sin?2+θ?-sin?π-θ?cos θ-sin θ1-tan θ1-2??-2. 答案:B 5124.解析:因为cos α=-13,且α是第三象限角,所以sin α=-13,5π?π???12cos?α+2?=cos?α+2?=-sin α=13. ????12答案:13 tan?-α??-sin α?cos?-α?5.解析:原式= ?-cos α?·sin α
2020-2021学年高中数学新教材必修第一册(人教A版)课时练习 5.3.2 诱导公式(二)(含解析)
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