Question

Решите подробно.....

Answer 1

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

$sinx=pm frac{tgx}{sqrt{1+tg^2x}}\\0<arctg3<frac{pi}{2}; ; to ; ; sin(arctg3)>0$

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

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