已知cos(π/4+x)=4/5,x∈(-π/2,-π/4),求(sin2x-2sin²x/1+tanx)的值
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已知cos(π/4+x)=4/5,x∈(-π/2,-π/4),求(sin2x-2sin²x/1+tanx)的值.
![已知cos(π/4+x)=4/5,x∈(-π/2,-π/4),求(sin2x-2sin²x/1+tanx)的值](/uploads/image/z/16912061-53-1.jpg?t=%E5%B7%B2%E7%9F%A5cos%EF%BC%88%CF%80%2F4%2Bx%EF%BC%89%3D4%2F5%2Cx%E2%88%88%EF%BC%88-%CF%80%2F2%2C-%CF%80%2F4%EF%BC%89%2C%E6%B1%82%EF%BC%88sin2x-2sin%26sup2%3Bx%2F1%2Btanx%EF%BC%89%E7%9A%84%E5%80%BC)
∵cos(π/4+x)=4/5,
∴cosx-sinx=(4/5)√2
两边平方得,cos²x+sin²x-2sinxcosx=32/25
则,2sinxcosx=-7/25
∴(cosx+sinx)²=1+2sinxcosx=18/25
∵x∈(-π/2,-π/4)
∴cosx+sinx=(-3/5)√2
解得sinx=(-7/10)√2
cosx=(√2)/10
∴(sin2x-2sin²x)/(1+tanx),注:下次出题时要加几个括号,以免题目表述不清,这除是斜的,不是横的,容易误解.
=(2sinxcosx-2sin²x)/(1+sinx/cosx)
=(-7/25-49/25)/(1-7)
=28/75
∴cosx-sinx=(4/5)√2
两边平方得,cos²x+sin²x-2sinxcosx=32/25
则,2sinxcosx=-7/25
∴(cosx+sinx)²=1+2sinxcosx=18/25
∵x∈(-π/2,-π/4)
∴cosx+sinx=(-3/5)√2
解得sinx=(-7/10)√2
cosx=(√2)/10
∴(sin2x-2sin²x)/(1+tanx),注:下次出题时要加几个括号,以免题目表述不清,这除是斜的,不是横的,容易误解.
=(2sinxcosx-2sin²x)/(1+sinx/cosx)
=(-7/25-49/25)/(1-7)
=28/75
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