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sin2xCos3x怎样才能变成sin5x%sinx

sin2xcos3x =1/2[sin(2x+3x)+sin(2x-3x)] =1/2(sin5x-sinx)

积化和差公式: sinαcosβ=1/2[sin(α+β)+sin(α-β)] 本题中,将3x看做α,2x看做β: sin3xcos2x = 1/2[sin(3x+2x)+sin(3x-2x)] = 1/2(sin5x+sinx)

使用积化和差公式一步到位 sinxcosy=(sin(x+y)+sin(x-y))/2

根据公式sinacosb=1/2(sin(a+b)+sin(a-b))得 sin2xcos3x=1/2(sin(2x+3x)+sin(2x-3x)) =1/2(sin5x+sin(-x)) =(sin5x-sinx)/2

三角函数的积化和差学过吗 cosα·sinβ=(1/2)[sin(α+β)-sin(α-β)]

cosx*cos2x*cos4x = 2 sinx*cosx*cos2x*cos4x / (2sinx) = sin2x * cos2x *cos4x /(2sinx) =......= sin8x / (8sinx) cos3x*cos5x =(1/2) ( cos8x +cos2x) 原式= (1/16) (1/sinx) [ sin8x cos8x + sin8xcos2x ] = (1/32) (1/sinx) [ sin16x + si...

积化和差公式sinαcosβ=1/2[sin(α+β)+sin(α-β)] ∴sin2xccos3x=1/2[sin(2x+3x)+sin(2x-3x)]=1/2(sin5x-sinx)

2、tan^2x=sin^2x/cos^2x=(1-cos^2x)/cos^2x...=∫tanxdtanx-∫(sinx/cosx)dx=(tan²x)/2-...(2x+3x)+sin(2x-3x))dx =1/2∫sin5xdx-1/...

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