Question

Докажите тождество!!!!!!!

Answer 1

Решение:

Рассмотрим левую часть: $sin^2(\frac{15\pi}{8}-2a)-cos^2(\frac{17\pi}{8}-2a)$

Вспоминаем формулы понижения степени: $sin^2(\frac{x}{2})=\frac{1-cosx}{2}$   ;   $cos^2(\frac{x}{2})=\frac{1+cosx}{2}$

$sin^2(\frac{15\pi}{8}-2a)-cos^2(\frac{15\pi}{8}-2a) = \frac{1-cos(\frac{15\pi}{4}-4a)}{2} - \frac{1+cos(\frac{17\pi}{4}-4a)}{2} = \frac{-cos(\frac{17\pi}{4}-4a)-cos(\frac{15\pi}{4}-4a)}{2}$

Вспоминаем формулу для косинуса разности: $cos(a-b)=cosacosb+sinasinb$

$\frac{-cos(\frac{17\pi}{4}-4a)-cos(\frac{15\pi}{4}-4a)}{2} = \frac{-(cos(\frac{17\pi}{4})cos(4a)+sin(\frac{17\pi}{4})sin(4a))-(cos(\frac{15\pi}{4})cos(4a)+sin(\frac{15\pi}{4})sin(4a))}{2}=\frac{-cos(\frac{17\pi}{4})cos(4a)-sin(\frac{17\pi}{4})sin(4a)-cos(\frac{15\pi}{4})cos(4a)-sin(\frac{15\pi}{4})sin(4a)}{2}$

Далее вычислим:

$cos(\frac{17\pi}{4})=cos(\frac{15\pi}{4})=\frac{\sqrt{2}}{2}$

$sin(\frac{17\pi}{4})=\frac{\sqrt{2}}{2}\\\\sin(\frac{15\pi}{4})=-\frac{\sqrt{2}}{2}$

Подставим:

$\frac{-cos(\frac{17\pi}{4})cos(4a)-sin(\frac{17\pi}{4})sin(4a)-cos(\frac{15\pi}{4})cos(4a)-sin(\frac{15\pi}{4})sin(4a)}{2}=\\\\ \frac{-\frac{\sqrt{2}}{2}cos4a-\frac{\sqrt{2}}{2}sin4a-\frac{\sqrt{2}}{2}cos4a+\frac{\sqrt{2}}{2}sin4a}{2}=\frac{-\sqrt{2}cos4a}{2}=\frac{-\cos4a}{\sqrt{2}}$ , что и требовалось доказать.

Answer 2

$Sin^{2}(\frac{15\pi }{8} -2\alpha)-Cos^{2}(\frac{17\pi }{8}-2\alpha)=Sin^{2}(2\pi-(\frac{\pi }{8}+2\alpha))-Cos^{2}(2\pi +(\frac{\pi }{8}-2\alpha))=Sin^{2}(\frac{\pi }{8}+2\alpha )-Cos^{2}(\frac{\pi }{8}-2\alpha)=(Sin(\frac{\pi }{8}+2\alpha)-Cos(\frac{\pi }{8}-2\alpha))*(Sin(\frac{\pi }{8}+2\alpha)+Cos(\frac{\pi }{8} -2\alpha))=(Sin(\frac{\pi }{8} +2\alpha)-Sin(\frac{3\pi }{8}+2\alpha))*(Sin(\frac{\pi }{8}+2\alpha)+Sin(\frac{3\pi }{8}+2\alpha)=$$2Sin(-\frac{\pi }{8})Cos(\frac{\pi }{4}+2\alpha)*2Sin(\frac{\pi }{4}+2\alpha)Cos(-\frac{\pi }{8})=(-2Sin\frac{\pi }{8}Cos\frac{\pi }{8})*(2Sin(\frac{\pi }{4}+2\alpha)Cos(\frac{\pi }{4}+2\alpha))=-Sin\frac{\pi }{4}*Sin(\frac{\pi }{2}+4\alpha)=-\frac{1}{\sqrt{2}}*Cos4\alpha=-\frac{Cos4\alpha}{\sqrt{2}}\\\\-\frac{Cos4\alpha}{\sqrt{2}}=-\frac{Cos4\alpha}{\sqrt{2}}$