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第五节 两角和与差的正弦、余弦和正切公式及二倍角公式
A组 基础题组
1.已知sinπ2+α=12,-π2<α<0,则cosα-π3的值是( )
A.12 B.23 C.-12 D.1
2.已知cosx-π6=-33,则cos x+cosx-π3=( )
A.-233 B.±233 C.-1 D.±1
3.已知sinα-π4=7210,cos 2α=725,则sin α=( )
A.45 B.-45 C.35 D.-35
4.(2016江西九校联考)已知锐角α,β满足sin α-cos α=16,tan α+tan β+3tan αtan β=3,则α,β的大小关系是( )
A.α<π4<β B.β<π4<α
C.π4<α<β D.π4<β<α
5.已知cos θ=-513,θ∈π,3π2,则sinθ-π6的值为 .
6.已知sinπ6-α=13,则cos2π3+α的值是 .
7.计算sin250°1+sin10°= .
8.已知α∈0,π2,tan α=12,求tan 2α和sin2α+π3的值.
9.已知α∈π2,π,且sinα2+cosα2=62.
(1)求cos α的值;
(2)若sin(α-β)=-35,β∈π2,π,求cos β的值.
B组 提升题组
10.cos π9·cos 2π9·cos-23π9=( )
A.-18 B.-116 C.116 D.18
11.定义运算a bc d=ad-bc.若cos α=17,sinα sinβcosα cosβ=3314,0<β<α<π2,则β等于( )
A.π12 B.π6 C.π4 D.π3
12.设α∈0,π2,β∈0,π2,且tan α=1+sinβcosβ,则( )
A.3α-β=π2 B.3α+β=π2
C.2α-β=π2 D.2α+β=π2
13.(2015广东,16,12分)已知tan α=2.
(1)求tanα+π4的值;
(2)求sin2αsin2α+sinαcosα-cos2α-1的值.
14.已知cosπ6+α·cosπ3-α=-14,α∈π3,π2.
(1)求sin 2α的值;
(2)求tan α-1tanα的值.
答案全解全析
A组 基础题组
1.C 由sinπ2+α=12得cos α=12,∵-π2<α<0,∴sin α=-32,∴cosα-π3=12cos α+32sin α=-12.
2.C ∵cosx-π6=-33,∴cos x+cosx-π3=cos x+cos xcosπ3+sin xsinπ3
=32cos x+32sin x=332cosx+12sinx
=3cosx-π6=3×-33=-1.
3.C 由sinα-π4=7210得sin α-cos α=75,①
由cos 2α=725得cos2α-sin2α=725,
所以(cos α-sin α)·(cos α+sin α)=725,②
由①②可得cos α+sin α=-15,③
由①③可得sin α=35.
4.B ∵α为锐角,sin α-cos α=16>0,∴α>π4.
又tan α+tan β+3tan αtan β=3,
∴tan(α+β)=tanα+tanβ1-tanαtanβ=3,
由题意知0<α+β<π,
∴α+β=π3,又α>π4,∴β<π4<α.
5.答案 5-12326
解析 由cos θ=-513,θ∈π,3π2得sin θ=-1-cos2θ=-1213,故sinθ-π6
=sin θcosπ6-cosθsinπ6=-1213×32--513×12=5-12326.
6.答案 -79
解析 ∵sinπ6-α=13,
∴cosπ3-2α=cos2π6-α
=1-2sin2π6-α=79,
∴cos2π3+α=cos2π3+2α
=cosπ-π3-2α=-cosπ3-2α=-79.
7.答案 12
解析 sin250°1+sin10°=1-cos100°2(1+sin10°)=1-cos(90°+10°)2(1+sin10°)=1+sin10°2(1+sin10°)=12.
8.解析 ∵tan α=12,
∴tan 2α=2tanα1-tan2α=2×121-14=43,
且sinαcosα=12,即cos α=2sin α,
又sin2α+cos2α=1,
∴5sin2α=1,而α∈0,π2,
∴sin α=55,则cos α=255.
∴sin 2α=2sin αcos α=2×55×255=45,
cos 2α=cos2α-sin2α=45-15=35,
∴sin2α+π3=sin 2αcosπ3+cos 2αsinπ3=45×12+35×32=4+3310.
9.解析 (1)将sinα2+cosα2=62两边同时平方,得1+sin α=32,则sin α=12.
又π2<α<π,所以cos α=-1-sin2α=-32.
(2)因为π2<α<π,π2<β<π,所以-π2<α-β<π2.
所以由sin(α-β)=-35,得cos(α-β)=45,
所以cos β=cos[α-(α-β)]
=cos αcos(α-β)+sin αsin(α-β)
=-32×45+12×-35=-43+310.
B组 提升题组
10.A cos π9·cos 2π9·cos-23π9
=cos 20°·cos 40°·cos 100°
=-cos 20°·cos 40°·cos 80°
=-sin20°·cos20°·cos40°·cos80°sin20°
=-12sin40°·cos40°·cos80°sin20°
=-14sin80°·cos80°sin20°
=-18sin160°sin20°=-18sin20°sin20°=-18.
11.D 依题意得sin α·cos β-cos α·sin β=sin(α-β)=3314.∵0<β<α<π2,∴0<α-β<π2,∴cos(α-β)=1314.
∵cos α=17,0<α<π2,
∴sin α=437.∴sin β=sin[α-(α-β)]=sin α·cos(α-β)-cos α·sin(α-β)=437×1314-17×3314=32,∴β=π3.故选D.
12.C 由tan α=1+sinβcosβ,得sinαcosα=1+sinβcosβ,即sin αcos β=cos α+sin βcos α,所以sin(α-β)=cos α,
又cos α=sinπ2-α,所以sin(α-β)=sinπ2-α,又因为α∈0,π2,β∈0,π2,所以-π2<α-β<π2,0<π2-α<π2,因此α-β=π2-α,所以2α-β=π2,故选C.
13.解析 (1)因为tan α=2,
所以tanα+π4=tanα+tanπ41-tanα·tanπ4
=2+11-2×1=-3.
(2)因为tan α=2,
所以sin2αsin2α+sinαcosα-cos2α-1
=2sinαcosαsin2α+sinαcosα-(cos2α-sin2α)-(sin2α+cos2α)
=2sinαcosαsin2α+sinαcosα-2cos2α
=2tanαtan2α+tanα-2=2×222+2-2=1.
14.解析 (1)cosπ6+α·cosπ3-α=cosπ6+α·sinπ6+α=12sin2α+π3=-14,
即sin2α+π3=-12.
∵α∈π3,π2,∴2α+π3∈π,4π3,
∴cos2α+π3=-32,
∴sin 2α=sin2α+π3-π3=sin2α+π3cosπ3-cos2α+π3sinπ3=12.
(2)∵α∈π3,π2,∴2α∈2π3,π,
又由(1)知sin 2α=12,∴cos 2α=-32.
∴tan α-1tanα=sinαcosα-cosαsinα=sin2α-cos2αsinαcosα=-2cos2αsin2α=(-2)×-3212=23.