Question

Помогите решить 3 задание

Answer 1

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

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

Ответ: 2√6/3.

Answer 2

$\frac{1}{Sin\alpha+Sin3\alpha}+\frac{1}{Sin3\alpha+Sin5\alpha}=\frac{1}{2Sin\frac{\alpha+3\alpha}{2}Cos\frac{\alpha-3\alpha}{2}}+\frac{1}{2Sin\frac{3\alpha+5\alpha}{2} Cos\frac{3\alpha-5\alpha}{2} }=\frac{1}{2Sin2\alpha Cos\alpha}+\frac{1}{2Sin4\alpha Cos\alpha}=\frac{Sin4\alpha+Sin2\alpha}{2Sin2\alpha Sin4\alpha Cos\alpha}=\frac{2Sin3\alpha Cos\alpha}{2Sin2\alpha Sin4\alpha Cos\alpha}=\frac{Sin3\alpha }{Sin2\alpha Sin4\alpha}$

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