7.3 正切函数的诱导公式
1.已知角α终边上有一点P(5n,4n)(n≠0),则tan(180°-α)的值是( )
A.-4/5 B.-3/5 C.±3/5 D.±4/5
解析∵角α终边上有一点P(5n,4n)(n≠0),∴tan α=4/5,
∴tan(180°-α)=-tan α=-4/5.
答案A
2.给出下列各函数值,其中符号为负的是( )
A.sin(-1 000°) B.cos(-2 200°)
C.tan(-10) D.(sin 7π/10 cos" " π)/(tan 17π/9)
解析sin(-1 000°)=sin(-3×360°+80°)=sin 80°>0;
cos(-2 200°)=cos 2 200°=cos(6×360°+40°)=cos 40°>0;
tan(-10)=-tan 10<0;
sin7π/10>0,cos π=-1<0,tan17π/9=tan8π/9<0,
故(sin 7π/10 cos" " π)/(tan 17π/9)>0.
答案C
3.已知tan(π+α)+1/(tan"(" 3π+α")" )=2,则tan(π-α)=( )
A.2 B.-2 C.1 D.-1
解析由已知可得tan α+1/tanα=2,解得tan α=1.
于是tan(π-α)=-tan α=-1.
答案D
4.导学号93774027设a=sin 33°,b=cos 55°,c=tan 35°,则( )
A.a>b>c B.b>c>a
C.c>b>a D.c>a>b
解析∵b=cos 55°=sin 35°>sin 33°=a,∴b>a.
又∵c=tan 35°=sin35"°" /cos35"°" >sin 35°=cos 55°=b,
∴c>b.∴c>b>a.故选C.
答案C
5.sin4π/3·cos5π/6·tan("-" 4π/3)的值是( )
A.-(3√3)/4 B.(3√3)/4
C.-√3/4 D.√3/4
解析原式=sin(π+π/3)·cos(π"-" π/6)·tan("-" π"-" π/3)=-sinπ/3·("-" cos π/6)·("-" tan π/3)=-√3/2×("-" √3/2)×(-√3)=-(3√3)/4.故选A.
答案A
6.tan("-" 23π/3)= .
解析tan("-" 23π/3)=-tan23π/3=-tan(7π+2π/3)=-tan2π/3=tanπ/3=√3.
答案√3
7.已知tan(π-x)=1/3,则tan(x-3π)= .
解析由tan(π-x)=1/3知,tan x=-1/3,
故tan(x-3π)=-tan(3π-x)=tan x=-1/3.
答案-1/3