Question

Помогите решить 3 задание

Answer 1

$frac{1}{sin{alpha }+sin{3alpha }} +frac{1}{sin{3alpha }+sin{5alpha }} =\frac{1}{2sin{2alpha }*cos{alpha }} +frac{1}{2sin{4alpha }*cos{alpha }} =\ frac{sin{4alpha }+sin{2alpha }}{2sin{2alpha }*cos{alpha }*sin{4alpha }} =\frac{2sin{3alpha }*cos{alpha }}{2sin{2alpha }*sin{4alpha }*cos{alpha }} =\frac{sin{3alpha }}{sin{2alpha }*sin{4alpha }}=\frac{sin{frac{pi}{4} }}{sin{frac{pi}{6} *sin{frac{pi}{3} }}} =$

$frac{frac{sqrt{2} }{2} }{frac{1}{2} *frac{sqrt{3} }{2} } =\frac{2sqrt{2} }{sqrt{3} }=\frac{2sqrt{6} }{3}$

Ответ: 2√6/3.

Answer 2

$frac{1}{Sinalpha+Sin3alpha}+frac{1}{Sin3alpha+Sin5alpha}=frac{1}{2Sinfrac{alpha+3alpha}{2}Cosfrac{alpha-3alpha}{2}}+frac{1}{2Sinfrac{3alpha+5alpha}{2} Cosfrac{3alpha-5alpha}{2} }=frac{1}{2Sin2alpha Cosalpha}+frac{1}{2Sin4alpha Cosalpha}=frac{Sin4alpha+Sin2alpha}{2Sin2alpha Sin4alpha Cosalpha}=frac{2Sin3alpha Cosalpha}{2Sin2alpha Sin4alpha Cosalpha}=frac{Sin3alpha }{Sin2alpha Sin4alpha}$

$alpha=frac{pi }{12}\\frac{Sin(3*frac{pi }{12}) }{Sin(2*frac{pi }{12})*Sin(4*frac{pi }{12})}=frac{Sinfrac{pi }{4} }{Sinfrac{pi }{6}*Sinfrac{pi }{3}}=frac{frac{sqrt{2} }{2} }{frac{1}{2}*frac{sqrt{3} }{2}}=frac{sqrt{2}*4 }{2*sqrt{3} }=frac{2sqrt{2} }{sqrt{3} }=frac{2sqrt{6} }{3}$