求证sin(x+3/1π)-根号3cos(3/2π-x)=2sin(x-3/1π)=0
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求证sin(x+3/1π)-根号3cos(3/2π-x)=2sin(x-3/1π)=0
对于这个 使用三角函数展开
sin(x+y)=sinx *cosy + cos x* siny
代入
sin(x+3/1π)=sin x cos(3/1π)+cosx sin(3/1π)=(1/2)*sinx+根号3*cosx
根号3cos(3/2π-x)=(-根号3)/2 * cos x+(根号3)/2 * sinx
sin(x+3/1π)-根号3cos(3/2π-x)=-sinx+根号3* cosx=-2(sin(x-π/3))
-2(sin(x-π/3))+2(sin(x-π/3))=0
sin(x+y)=sinx *cosy + cos x* siny
代入
sin(x+3/1π)=sin x cos(3/1π)+cosx sin(3/1π)=(1/2)*sinx+根号3*cosx
根号3cos(3/2π-x)=(-根号3)/2 * cos x+(根号3)/2 * sinx
sin(x+3/1π)-根号3cos(3/2π-x)=-sinx+根号3* cosx=-2(sin(x-π/3))
-2(sin(x-π/3))+2(sin(x-π/3))=0
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