2015届高考数学一轮总复习45简单的三角恒等变换
基础巩固强化一、选择题π1θ1.文设θπ,且cosθ=,那么si
的值为252A105155B.-D155105
C.-
答案Dπ1解析∵θπ,∴cosθ0,∴cosθ=-25πθπθ∵,∴si
0,4222θθ1-cosθ3又cosθ=1-2si
2,∴si
2==,2225θ15∴si
=25π理已知x∈,π,cos2x=a,则cosx=2AC1-a21+a2B.-D.-1-a21+a2
答案D解析a=cos2x=2cos2x-1,π∵x∈,π,∴cosx0,∴cosx=-2a+12
π32.2013山西诊断已知si
+θ=,则cosπ-2θ=2512A257C.-25答案D解析D12B.-257D25
π337依题意得si
θ+=cosθ=,cosπ-2θ=-cos2θ=1-2cos2θ=1-2×2=,选25525
ACAC3.文在△ABC中,A、B、C成等差数列,则ta
+ta
+3ta
ta
的值是2222A.±3
1
B.-3
fC3答案C
D
33
解析∵A、B、C成等差数列,∴2B=A+C,π2π又A+B+C=π,∴B=,A+C=,33ACAC∴ta
+ta
+3ta
ta
2222ACACAC+1-ta
ta
+3ta
ta
=ta
222222=3,故选C理2013兰州名校检测在斜三角形ABC中,si
A=-2cosBcosC,且ta
Bta
C=1-2,则角A的值为πA4πC2答案A解析由题意知,si
A=-2cosBcosC=si
B+C=si
BcosC+cosBsi
C,在等式-2πB33πD4
cosBcosC=si
BcosC+cosBsi
C两边同除以cosBcosC得ta
B+ta
C=-2,ta
B+ta
Cπ又ta
B+C==-1=-ta
A,即ta
A=1,所以A=41-ta
Bta
C14.文若cosx+ycosx-y=,则cos2x-si
2y等于31A.-32C.-3答案B解析∵cosx+ycosx-y=cosxcosy-si
xsi
ycosxcosy+si
xsi
y=cos2xcos2y-si
2xsi
2y1B32D3
=cos2x1-si
2y-1-cos2xsi
2y=cos2x-cos2xsi
2y-si
2y+cos2xsi
2y=cos2x-si
2y,∴选B22理已知si
x-si
y=-,cosx-cosy=,且x、y为锐角,则si
x+y的值是33A.11C3答案A解析两式相加得si
x+cosx=si
y+cosy,ππ∴si
x+4=si
y+4,∵x、y为锐角,且si
x-si
y0,∴xy,
2
B.-11D2
fπππ∴x+=π-y+4,∴x+y=2,4∴si
x+y=1π45.已知α∈-2,0,cosα=5,则ta
2α等于24A.-77C.-24答案Aπ4解析∵-α0,cosα=,253si
α3∴si
α=-1-cos2α=-,∴ta
α==-,5cosα4∴ta
2α=2ta
α242=-,故选A71-ta
α24B77D24
πβ3α16.若α、β∈0,,cosα-=,si
-β=-,则cosα+βr