.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α
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.[1/cos(-α)+cos(180°+α)]/[1/sin(540°-α)+sin(360°-α)]=tan^3α.[1/cos(-α)+cos(180°+α)]/[1/sin(540°-α)+s
.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α
.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α
.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α
cos(-α)=cosα,
cos(180°+ α)= -cosα
sin(540°-α)=sinα
sin(360°-α)= -sinα
所以
原式左边
=(1/cosα -cosα) / (1/sinα -sinα)
=(1-cos²α)/cosα / (1-sin²α)/sinα
=(sin²α/cosα) / (cos²α)/cosα
=(sinα/cosα)^3
=tan^3 α
这样就得到了答案
怎样化简cosαcos(60°-α)+1
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cos
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已知cos(75°+α)=1/3,-180°
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Cos(180°-α)=( )
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.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α
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请问:当α为锐角时,为什么有cos(180°+α)=-cosα?
cosα+cos(180-α)=?推导过程