已知sin(Π/6-α)=2cos(α-4Π),求cos(Π/2-α)+5sin(Π/2+α)分之2cos(Π/2+α)
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已知sin(Π/6-α)=2cos(α-4Π),求cos(Π/2-α)+5sin(Π/2+α)分之2cos(Π/2+α)-sin(-α)
sin(π/6-α)=2cos(α-4π)
sin(π/6-α)=2cosα
sinπ/6cosα-cosπ/6sinα-=2cosα
1/2cosα-√3/2sinα-=2cosα
3/2cosα=-√3/2sinα
3cosα=-√3sinα
sinα=-√3cosα
[2cos(π/2+α)-sin(-α)]/[cos(π/2-α)+5sin(π/2+α)]
=(-2sinα+sinα)/(sinα+5cosα)
=(-sinα)/(sinα+5cosα)
=(-√3cosα)/(-√3cosα+5cosα)
=-√3/(-√3+5)
=-√3/(5-√3)
=-√3(5+√3)/(5-√3)(5+√3)
=-(5√3+3)/22
再问: 我前面那块打错了,应该是(3Π+α)=2cos(α-4Π)应该怎么做呀?
再答: sin(3Π+α)=2cos(α-4Π)是这样吗?
再问: 是的
再答: sin(3π+α)=2cos(α-4π) sin(π+α)=2cosα -sinα=2cosα sinα=-2cosα [2cos(π/2+α)-sin(-α)]/[cos(π/2-α)+5sin(π/2+α)] =(-2sinα+sinα)/(sinα+5cosα) =(-sinα)/(sinα+5cosα) =[-(-2cosα)]/(-2cosα+5cosα) =2cosα/(3cosα) =-2/3
sin(π/6-α)=2cosα
sinπ/6cosα-cosπ/6sinα-=2cosα
1/2cosα-√3/2sinα-=2cosα
3/2cosα=-√3/2sinα
3cosα=-√3sinα
sinα=-√3cosα
[2cos(π/2+α)-sin(-α)]/[cos(π/2-α)+5sin(π/2+α)]
=(-2sinα+sinα)/(sinα+5cosα)
=(-sinα)/(sinα+5cosα)
=(-√3cosα)/(-√3cosα+5cosα)
=-√3/(-√3+5)
=-√3/(5-√3)
=-√3(5+√3)/(5-√3)(5+√3)
=-(5√3+3)/22
再问: 我前面那块打错了,应该是(3Π+α)=2cos(α-4Π)应该怎么做呀?
再答: sin(3Π+α)=2cos(α-4Π)是这样吗?
再问: 是的
再答: sin(3π+α)=2cos(α-4π) sin(π+α)=2cosα -sinα=2cosα sinα=-2cosα [2cos(π/2+α)-sin(-α)]/[cos(π/2-α)+5sin(π/2+α)] =(-2sinα+sinα)/(sinα+5cosα) =(-sinα)/(sinα+5cosα) =[-(-2cosα)]/(-2cosα+5cosα) =2cosα/(3cosα) =-2/3
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