Question

Упростите выражение с тригонометрическими функциями

Answer 1

Ответ:   c) .

$\displaystyle \frac{sina+cosa}{cosa-sina}=\frac{sina+sin(\frac{\pi}{2}-a)}{sin(\frac{\pi}{2}-a)-sina}=\frac{2\cdot sin\frac{\pi}{4}\cdot cos(a-\frac{\pi}{4})}{2\cdot sin(\frac{\pi}{4}-a)\cdot cos\frac{\pi}{4}}=\\\\\\=tg\frac{\pi}{4}\cdot \frac{cos(\frac{\pi}{4}-a)}{sin(\frac{\pi}{4}-a)}=1\cdot ctg\Big(\frac{\pi}{4}-a\Big)=tg\Big(\frac{\pi}{2}-\Big(\frac{\pi}{4}-a\Big)\Big)=tg\Big(\frac{\pi}{4}+a\Big)$

$\star \ \ sin\Big(\dfrac{\pi}{2}-x\Big)=cosx\ \ ,\ \ \ ctg\Big(\dfrac{\pi}{2}-x\Big)=tgx\ \ \star$

$\star \ \ sina\pm sin\beta =2\cdot sin\dfrac{a\pm \beta }{2}\cdot cos\dfrac{a\mp \beta }{2}\ \ \star$

Answer 2

Ещё один способ :

$\dfrac{Sin\alpha+Cos\alpha}{Cos\alpha-Sin\alpha}=\dfrac{\sqrt{2}\Big(\dfrac{1}{\sqrt{2} }\cdot Sin\alpha+\dfrac{1}{\sqrt{2} }\cdot Cos\alpha\Big)}{\sqrt{2}\Big(\dfrac{1}{\sqrt{2} }\cdot Cos\alpha-\dfrac{1}{\sqrt{2} }\cdot Sin\alpha\Big) } =\\\\=\dfrac{Cos\dfrac{\pi }{4}Sin\alpha+Sin\dfrac{\pi }{4}Cos\alpha}{Cos\dfrac{\pi }{4}Cos\alpha-Sin\dfrac{\pi }{4}Sin\alpha}=\dfrac{Sin\Big(\frac{\pi }{4}+\alpha\Big) }{Cos\Big(\dfrac{\pi }{4}+\alpha\Big)} =\boxed{tg\Big(\dfrac{\pi }{4}+\alpha\Big)}$