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)
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 + 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)
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^3
化简sin(α+180°)cos(-α)sin(-α-180°)
①sin(α+180°)cos(-α)sin(-α-180°)和②sin³(-α)cos(2π+α)tan(-
若sin(α-360°)-cos(180°-α)=m,则sin(180°+α)·cos(180°-α)等于
求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)
化简cos(90°-α)/sin(270°-α)*sin(360°-α)*sin(180°-α)
化简cos^2(-α)-[tan(360°+α)/sin(-α)]
化简,cos²(-α)-tan(360°+α)/sin(-α)
已知sin(360°+α)-cos(180°-α)=m
已知cos(180°+α)=-3/5,则sin(360°-α)的值等于