[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
[sin(45°+θ)]/[cos(45°-θ)]-[cos(θ+60°)]/[sin(θ+30°)]=?
sin(θ+75°)+cos(θ+45°)-3cos(θ+15°
sinΘ-2cosΘ=0,Θ属于(0.,90°),求sinΘ、cosΘ的值.
2cos 30°-2sin 60°·cos 45°=?
(cos 45°-sin 30°)/(cos 45°+sin 30°)
化简:cos平方(θ+15°)+sin平方(θ-15°)+sin(θ+180°)cos(θ-180°)
cos tan sin 30° 45° 60°是多少
化简[ sec(-θ)+cos(-θ+180° ) ] / [ cos(180 °-θ)+sin(360 °-θ ) ]
(cosθ-sinθ)/(cosθ+sinθ)等于多少
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
sin(cosθ)是不是sin乘以cosθ呀
若sinθ*cosθ