Question

Как решать такое и подобные?

Answer 1

$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\\\\\cos\left(\frac\pi4+\alpha\right)=\cos\frac\pi4\cos\alpha-\sin\frac\pi4\sin\alpha=\frac1{\sqrt2}\cos\alpha-\frac1{\sqrt2}\sin\alpha=\frac1{\sqrt2}(\cos\alpha-\sin\alpha)\\\\\sin\alpha=-0,6\\\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}\\\cos\alpha=\sqrt{1-(-0,6)^2}=\sqrt{1-0,36}=\sqrt{0,64}=\pm0,8\\\frac{3\pi}2<\alpha<2\pi\Rightarrow \cos\alpha>0\\\cos\alpha=0,8$

$\cos\left(\frac\pi4+\alpha\right)=\frac1{\sqrt2}(0,8-(-0,6))=\frac{1,4}{\sqrt2}\approx1$