=-tan15°=-tan(45°-30°)=.
答案:
5.化简:-2cos(α-β).
解:-2cos(α-β)
6.已知cos(θ-α)=a,sin(θ-β)=b,求证:cos2(α-β)=a2+b2-2absin(α-β).
证明:由cos(θ-α)=a得cosθcosα+sinθsinα=a, ①
由sin(θ-β)=b得sinθcosβ-cosθsinβ=b, ②
①×sinβ+②×cosα得sinθcos(α-β)=asinβ+bcosα, ③
①×cosβ-②×sinα得cosθcos(α-β)=acosβ-bsinα, ④
③2+④2得cos2(α-β)=a2+b2+2ab(sinβcosα-cosβsinα)=a2+b2-2absin(α-β),
∴结论成立.
30分钟训练(巩固类训练,可用于课后)
1.(1+tan17°)(1+tan18°)(1+tan27°)(1+tan28°)的值是( )
A.2 B.4 C.8 D.16
解析:tan(α+β)=,当α+β=45°时,tanα+tanβ=1-tanαtanβ,
∴tanα+tanβ+tanαtanβ+1=2.
∴(1+tanα)(1+tanβ)=2.
∴(1+tan17°)(1+tan18°)=2,(1+tan27°)(1+tan28°)=2.
答案:B
2.y=3sin(x+10°)+5sin(x+70°)的最大值是( )
A. B. C.7 D.8
解析:y=3sin(x+10°)+5sin(x+70°)
=3sin(x+10°)+5sin(x+10°+60°)
=3sin(x+10°)+5sin(x+10°)cos60°+5cos(x+10°)sin60°
= sin(x+10°)+cos(x+10°),
∴y的最大值为()2+()2=7.