Question

(sinx+xsin3x+sin 5x）/(cosx+cos3x+cos5x)怎么化简？

Answer 1

我们可以使用三倍角公式来化简分母和分子，即：
cos3x = 4cos^3 x - 3cos x cos5x = 16cos^5 x - 20cos^3 x + 5cos x
sin3x = 3sin x - 4sin^3 x sin5x = 5sin x - 20sin^3 x + 16sin^5 x
将这些公式代入原式，得到：
(sinx+xsin3x+sin5x)/(cosx+cos3x+cos5x) = (sinx+x(3sinx-4sin^3x)+5sinx-20sin^3x+16sin^5x)/(cosx+4cos^3x-3cosx+16cos^5x-20cos^3x+5cosx) = (22sinx-20sin^3x+16sin^5x)/(16cos^5x+4cos^3x+2cosx)
然后，我们可以使用双角公式来进一步化简，即：
sin^2 x = (1 - cos 2x)/2 sin^4 x = (1 - cos 2x)^2/4 sin^5 x = sin^4 x * sin x = (1 - cos 2x)^2/4 * sin x
将这些公式代入上式，得到：
(22sinx-20sin^3x+16sin^5x)/(16cos^5x+4cos^3x+2cosx) = (22sinx-20sin^3x+16(1 - cos 2x)^2/4 * sin x)/(16cos^5x+4cos^3x+2cosx) = (22sinx-20sin^3x+(4 - 8cos 2x + 4cos^2 2x) * sin x)/(16cos^5x+4cos^3x+2cosx) = (22sinx-20sin^3x+4sin x - 8cos 2x * sin x + 4cos^2 2x * sin x)/(16cos^5x+4cos^3x+2cosx) = (26sinx-20sin^3x-8cos 2x * sin x + 4cos^2 2x * sin x)/(16cos^5x+4cos^3x+2cosx)
因此，原式可以化简为 (26sinx-20sin^3x-8cos 2x * sin x + 4cos^2 2x * sin x)/(16cos^5x+4cos^3x+2cosx)。