Question

даю 100 баллов!!!!!!!!!!!!!помогите пожалуйста!​

Answer 1

$\star \; \; sin^4x+cos^4x=(sin^4x+cos^4x+2sin^2x\, cos^2x)-2sin^2x\, cos^2x=\\\\=(sin^2x+cos^2x)^2-2\, sin^2x\, cos^2x=1^2-2\, (sinx\, cosx)^2=\\\\=1-2\cdot (\frac{1}{2}sin2x)^2=1-\frac{1}{2}\, sin^22x\; \; \star \\\\\\sin^4x+cos^4x=sin^22x-\frac{1}{2}\\\\1-\frac{1}{2}\, sin^22x=sin^22x-\frac{1}{2}\\\\\frac{3}{2}\, sin^22x=\frac{3}{2}\\\\sin^22x=1\\\\\frac{1-cos4x}{2}=1\\\\1-cos4x=2\\\\cos4x=-1\\\\4x=\pi +2\pi n\; ,\; n\in Z\\\\x=\frac{\pi}{4}+\frac{\pi n}{2}\; ,\; n\in Z$

$Formyla:\; \; \; sin^2a=\frac{1-cos2a}{2}\; .$