若(sinα+cosα)/(sinα-cosα)=2,求sin(α-5π)sin(3π/2-α)的值

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若(sinα+cosα)/(sinα-cosα)=2,求sin(α-5π)sin(3π/2-α)的值
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若(sinα+cosα)/(sinα-cosα)=2,求sin(α-5π)sin(3π/2-α)的值
若(sinα+cosα)/(sinα-cosα)=2,求sin(α-5π)sin(3π/2-α)的值

若(sinα+cosα)/(sinα-cosα)=2,求sin(α-5π)sin(3π/2-α)的值
∵(sinα+cosα)/(sinα-cosα)=2
==>sinα+cosα=2(sinα-cosα)
==>sinα=3cosα
==>(3cosα)^2+(cosα)^2=1
==>10(cosα)^2=1
∴(cosα)^2=1/10
故sin(α-5π)sin(3π/2-α)
=-sinα(-cosα)
=sinαcosα
=(3cosα)cosα
=3(cosα)^2
=3(1/10)
=3/10.

用合分比定理,得
(sinα+cosα)/(sinα-cosα)=2
→sinα/cosα=3
→tanα=3.
∴sin(α-5π)sin(3π/2-α)
=(-sinα)·(-cosα)
=sinαcosα/(sin²α+cos²α)
=tanα/(1+tan²α)
=3/(1+3²)
=3/10.