2cosθ^4+5cosθ^2-7=asinθ^4+bsinθ^2+c如果等式成立,求a,b,c

1+cos(2θ)+cos(4θ)+cos(6θ)=4cosθcos(2θ)cos(3θ)证明相等. 已知sinθsinβ=-4/7,则cosθcosβ∈已知tan(π-θ)=3,求下式的值 (5xin^3θ+cosθ)/(2cos^3θ+sin^2θcosθ) cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值 cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值 5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β cos²θ-2cos-3/cos²θ+cosθ为什么等于cosθ-3/cosθ cosθ>0,cos(θ/2) 已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcosθ-3cos^2θ) 为什么cos(θ-2π)=cosθ, 若sinθ+sin^2θ=1,则cos^2θ+cos^4θ+cos^6θ 化简cosθ+cos(θ+2π/3)+cos(θ+4π/3) cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4 cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4 cos^2 π/8+cos ^2 3π/8+ cos^2 5π/8+cos^2 7π/8=? 已知sinΘ+cosΘ=2sinα,sinΘ*cosΘ=sin²β,求证：4cos²2α=cos²2β 已知sinθ+cosθ=2sinα,sinθ*cosθ=sin²β,求证：4cos²2α=cos²2β 已知sinθ+cosθ=2sinx,sinθcosθ=sin²y,求证：4cos²2x=cos²2y 1,已知sinθ-cosθ=-1/5求（1)sinθcosθ（2）sin^4θ+cos^4θ2,证明下列恒等式（1）2cos²θ+sin^4θ=cos^θ+1(2)sin^4θ+sin²θcos²θ+cos²θ=1