Решите подробно.....

$2\sqrt{10}\cdot sin(\underbrace {\frac{1}{2}arctg3}_{ \alpha })\cdot cos(\underbrace {\frac{1}{2}arctg3}_{ \alpha })=\Big [2\cdot sin \alpha \cdot cos\alpha =sin2 \alpha \Big ]=\\\\=\sqrt{10}\cdot sin(arctg3)\\\\\\sin^2x+cos^2x=1\; |:sin^2x\ne 0\\\\1+ctg^2x=\frac{1}{sin^2x}\; \; \; \to \; \; \; sin^2x=\frac{1}{1+ctg^2x}=\frac{1}{1+\frac{1}{tg^2x}}=\frac{tg^2x}{1+tg^2x}\; \to$

0" alt="sinx=\pm \frac{tgx}{\sqrt{1+tg^2x}}\\\\00" align="absmiddle" class="latex-formula">

$sin(arctg3)=\frac{tg(arctg3)}{\sqrt{1+tg^2(arctg3)}}=[tg(arctgx)=x]=\frac{3}{\sqrt{1+3^2}}=\frac{3}{\sqrt{10}}$

$\sqrt{10}\cdot sin(arctg3)=\sqrt{10}\cdot \frac{3}{\sqrt{10}}=3$