Question

Тригонометрия
10 класс

Помогите доказать тождества(с решением)

Answer 1

4)

$1)\frac{Sin4\alpha}{1+Cos4\alpha}*\frac{Cos2\alpha}{1+Cos2\alpha}=\frac{2Sin2\alpha Cos2\alpha}{2Cos^{2}2\alpha}*\frac{Cos2\alpha}{2Cos^{2} \alpha}=\frac{Sin2\alpha*Cos2\alpha}{Cos2\alpha*2Cos^{2}\alpha}=\frac{2Sin\alpha Cos\alpha}{2Cos^{2}\alpha}=\frac{Sin\alpha}{Cos\alpha}=tg\alpha\\\\2)Ctg(\frac{3\pi }{2}-\alpha)=tg\alpha$

tgα = tgα

Что и требовалось доказать

5)

$\frac{tg(\pi+2\alpha)*Ctg(\frac{3\pi }{2}+\alpha)}{tg2\alpha-tg\alpha}+2Cos(\frac{\pi }{4}-\alpha)Cos(\frac{\pi }{4}+\alpha)=\frac{tg2\alpha*(-tg\alpha)}{tg2\alpha-tg\alpha}+2*\frac{1}{2}(Cos(\frac{\pi }{4}-\alpha-\frac{\pi }{4}-\alpha)+Cos(\frac{\pi }{4}-\alpha+\frac{\pi }{4}+\alpha))=-\frac{tg2\alpha*tg\alpha}{tg2\alpha-tg\alpha}+Cos2\alpha+Cos\frac{\pi }{2}=$$Cos2\alpha-\frac{\frac{Sin2\alpha }{Cos2\alpha}*\frac{Sin\alpha}{Cos\alpha}}{\frac{Sin2\alpha }{Cos2\alpha}-\frac{Sin\alpha }{Cos\alpha}}=Cos2\alpha-\frac{2Sin^{2}\alpha}{Cos2\alpha}*\frac{Cos2\alpha*Cos\alpha}{Sin2\alpha Cos\alpha-Cos2\alpha *Sin\alpha} =Cos2\alpha-\frac{2Sin^{2}\alpha*Cos\alpha   }{Sin\alpha }=Cos2\alpha-Sin2\alpha$

$\sqrt{2}Sin(\frac{\pi }{4}-2\alpha)=\sqrt{2}(Sin\frac{\pi }{4}Cos2\alpha-Cos\frac{\pi }{4}Sin2\alpha=\sqrt{2}(\frac{\sqrt{2} }{2} Cos2\alpha-\frac{\sqrt{2} }{2}Sin2\alpha)}=\sqrt{2}*\frac{2}{2}(Cos2\alpha-Sin2\alpha)=Cos2\alpha-Sin2\alpha\\\\Cos2\alpha-Sin2\alpha=Cos2\alpha-Sin2\alpha$

Что и требовалось доказать