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青青子衿
发表于 2013-10-26 10:46
$f(x)=\sin x+\frac{1}{2}\sin2x+\frac{1}{3}\sin3x$
$f'(x)=\cos x+\cos2x+\cos3x=\cos x+2\cos^2x-1+4\cos^3x-3\cos x=2\cos^2x(2\cos x+1)-(2\cos x+1)=(2\cos^2x-1)(2\cos x+1)$
$f(x)=\sin x+\frac{1}{2}\sin2x+……+\frac{1}{n}\sin nx$
$f'(x)=\cos x+\cos2x+……+\cos nx=\frac{1}{2\sin\frac{x}{2}}(2\cos x\sin\frac{x}{2}+2\cos2x\sin\frac{x}{2}+……+2\cos nx\sin\frac{x}{2})=\frac{1}{2\sin\frac{x}{2}}[(\sin\frac{x}{2}-\sin\frac{3x}{2})+(\sin\frac{3x}{2}-\sin\frac{5x}{2})+……+(\sin\frac{2n-1}{2}x-\sin\frac{2n+1}{2}x)]=\frac{1}{2\sin\frac{x}{2}}(\sin\frac{x}{2}-\sin\frac{2n+1}{2}x)=\frac{\sin\frac{nx}{2}\cos\frac{n+1}{2}x}{sin\frac{x}{2}}$
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Tesla35
发表于 2013-10-13 14:19
