Question

Cos^2(3x)+cos^2(x)-cos(4x)*cos(2x)
подробно​

Answer 1

$\displaystyle cos^23x+cos^2x-cos4x\cdot cos2x=\\\\\\=\frac{1+cos6x}{2}+\frac{1+cos2x}{2}-\frac{1}{2}\cdot (cos8x+cos2x)=\\\\\\=\frac{1}{2}\cdot (1+cos6x+1+cos2x-cos8x-cos2x)=\frac{1}{2}\cdot (2+cos6x-cos8x)=\\\\\\=\frac{1}{2}\cdot (2+2\, sin7x\cdot sinx)=1+sin7x\cdot sinx$

$\star \ \ cos^2\alpha =\dfrac{1+cos2\alpha }{2}=\dfrac{1}{2}+\dfrac{1}{2}\, cos2\alpha \ \ \ \star$