已知cos(π/2+α)=sin(α-π/2) 求sin(π-α)-cos(π+α)/cos(5π/2 -α)+2sin
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已知cos(π/2+α)=sin(α-π/2) 求sin(π-α)-cos(π+α)/cos(5π/2 -α)+2sin(7π/2 -α)
如题 结果是-2 我算的结果是0
如题 结果是-2 我算的结果是0
先化简
cos(π/2+α) = cos(π/2 -(-α)) = sin(-α) =-sinα
sin(α-π/2) = sin(-(π/2-α)) = -sin(π/2-α)= - cosα
sin(π-α)= sinα
cos(π+α)= -cosα
cos(5π/2 -α)= cos(4π/2 + π/2 -α)= cos(π/2 -α) = sinα
sin(7π/2 -α)= sin(8π/2 -π/2-α) = sin(-π/2-α)= -sin(π/2-(-α) )= -cos(-α) = -cosa
由已知-sinα= - cosα
所以sinα=cosα
设sinα=cosα = p
sin(π-α)-cos(π+α)/cos(5π/2 -α)+2sin(7π/2 -α)
= sinα - (-cosα) / sinα + 2(-cosα)
= p +p / p -2p
= 2p/-p
= -2
cos(π/2+α) = cos(π/2 -(-α)) = sin(-α) =-sinα
sin(α-π/2) = sin(-(π/2-α)) = -sin(π/2-α)= - cosα
sin(π-α)= sinα
cos(π+α)= -cosα
cos(5π/2 -α)= cos(4π/2 + π/2 -α)= cos(π/2 -α) = sinα
sin(7π/2 -α)= sin(8π/2 -π/2-α) = sin(-π/2-α)= -sin(π/2-(-α) )= -cos(-α) = -cosa
由已知-sinα= - cosα
所以sinα=cosα
设sinα=cosα = p
sin(π-α)-cos(π+α)/cos(5π/2 -α)+2sin(7π/2 -α)
= sinα - (-cosα) / sinα + 2(-cosα)
= p +p / p -2p
= 2p/-p
= -2
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