Question

Решите пожалуйста, sin 2a, tg 2a, если tg a= корню из 2 , П

Answer 1

$\mathrm{tg}(a) = \sqrt{2}$

$\sin(2a) = 2\cdot\sin(a)\cos(a) = 2\cdot\frac{\sin(a)}{\cos(a)}\cdot\cos^2(a) = 2\cdot\mathrm{tg}(a)\cdot\cos^2(a)$

$\frac{1}{\cos^2(a)} = 1 + \mathrm{tg}^2(a)$

$\cos^2(a) = \frac{1}{1+\mathrm{tg}^2(a)}$

$\sin(2a) = 2\cdot\frac{\mathrm{tg}(a)}{1+\mathrm{tg}^2(a)}$

$\sin(2a) = 2\cdot\frac{\sqrt{2}}{1 + (\sqrt{2})^2} =$

$= 2\cdot\frac{\sqrt{2}}{1+2} = \frac{2\cdot\sqrt{2}}{3}$

$\mathrm{tg}(2a) = \frac{\sin(2a)}{\cos(2a)} =$

$= \frac{2\cdot\sin(a)\cdot\cos(a)}{\cos^2(a) - \sin^2(a)} =$

$= \frac{2\cdot\frac{\sin(a)}{\cos(a)}}{1 - \frac{\sin^2(a)}{\cos^2(a)}} =$

$= \frac{2\cdot\mathrm{tg}(a)}{1 - \mathrm{tg}^2(a)} =$

$= \frac{2\cdot\sqrt{2}}{1- (\sqrt{2})^2} = \frac{2\cdot\sqrt{2}}{1 - 2} =$

$= -2\cdot\sqrt{2}$