Question

数学必修四解三角函数

Answer 1

cos(α+β)cosβ+sin(α+β)sinβ=1/3
cos[(α+β)-β]=1/3
cosα=1/3
∵α∈(3π/2,2π)
∴sinα<0
sinα=-√(1-cos^2α)
=-2√2/3
cos(2α+π/4)
=cos2αcosπ/4-sin2αsinπ/4
=√2/2(cos2α-sin2α)
=√2/2(cos^2α-sin^2α-2sinαcosα)
=√2/2[(1/3)^2-(-2√2/3)^2-2(-2√2/3)(1/3)]
=√2/2[1/9-8/9+4√2/9]
=-7√2/18+4/9
=(8-7√2)/18

Answer 2

熟读书本定理，不要死记硬背，边做题边应用，借题理解，本人自身经验如此，绝非空话

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谢谢