Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)
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Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)Co
Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)
Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)
Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)
这道题是考三角函数的性质
cos(180 + a) = - cos a
sin ( a + 360) = sin a
sin(-a-180) = - sin ( a + 180 ) = sin a
cos( - 180 - a) = cos(180 + a) = - cos a
所以原式 = - cos a sin a / ( sin a * - cos a)
=1
Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)
sin(α+30°)-sin(α-30°)/cosα
化简cos(540°-α)sin(α-360°)/sin(-α+180°)cos(180°+α)
.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α
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