证明:(sinθ)^4+(sin2θ)^4/4+(sin4θ)^4/16+(sin8θ)^4/64=(sinθ)^2-(sin16θ)^/256

(sinθ)^4=(sinθ)^2（1-(cosθ)^2）=(sinθ)^2-(sinθcosθ)^2=(sinθ)^2-(sin2θ)^2/4
所以(sinθ)^4+(sin2θ)^4/4=(sinθ)^2-(sin2θ)^2/4+(sin2θ)^4/4=(sinθ)^2-(sin2θ)^2/4(1-(sin2θ)^2)=(sinθ)^2-(sin2θ)^2(cos2θ)^2)/4=(sinθ)^2-(sin4θ)^2/16
同理 一直加到(sin8θ)^4/64 同样的变换 结果等于(sinθ)^2-(sin16θ)^/256