证明:(sinθ)^4+(sin2θ)^4/4+(sin4θ)^4/16+(sin8θ)^4/64=(sinθ)^2-(
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证明:(sinθ)^4+(sin2θ)^4/4+(sin4θ)^4/16+(sin8θ)^4/64=(sinθ)^2-(sin16θ)^/256
![证明:(sinθ)^4+(sin2θ)^4/4+(sin4θ)^4/16+(sin8θ)^4/64=(sinθ)^2-(](/uploads/image/z/19429760-56-0.jpg?t=%E8%AF%81%E6%98%8E%3A%28sin%CE%B8%29%5E4%2B%28sin2%CE%B8%29%5E4%2F4%2B%28sin4%CE%B8%29%5E4%2F16%2B%28sin8%CE%B8%29%5E4%2F64%3D%28sin%CE%B8%29%5E2-%28)
(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
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
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