Question

Найдите sin 4 a, если tg a = -3 , - П/2

Answer 1

$\mathrm{tg}a = -3$

$\sin^4(a) = (\sin^2(a))^2 = (1-\cos^2(a))^2 = V$

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

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

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

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

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

$= (\frac{(-3)^2}{1+(-3)^2})^2 = (\frac{9}{10})^2 = \frac{81}{100} = 0{,}81$

$\sin(4a) = 2\sin(2a)\cos(2a) = W$

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

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

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

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

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

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

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

$= 2\cdot\frac{2\cdot(-3)}{1+(-3)^2}\cdot\frac{1 - (-3)^2}{1+(-3)^2} =$

$= 2\cdot\frac{-6}{1+9}\cdot\frac{1-9}{1+9} = \frac{2\cdot(-6)\cdot(-8)}{10\cdot 10} =$

$= \frac{2\cdot 6\cdot 8}{100} = \frac{96}{100} = 0{,}96$