Question

Вычислите: tg(arcsin(-1/3)+пи/2)

Answer 1

Формулы:

$\sin\left(\arcsin x\right)=x, \ -1\leq x\leq 1\\\cos\left(\arcsin x\right)=\sqrt{1-x^2} , \ -1\leq x\leq 1$

$\mathrm{tg}\left(\arcsin\left(-\dfrac{1}{3}\right)+\dfrac{\pi}{2}\right)=-\mathrm{ctg}\left(\arcsin\left(-\dfrac{1}{3}\right)\right)=\\=-\dfrac{\cos\left(\arcsin\left(-\frac{1}{3}\right)\right)}{\sin\left(\arcsin\left(-\frac{1}{3}\right)\right)} =-\dfrac{\sqrt{1-\left(-\frac{1}{3}\right)^2}}{-\frac{1}{3}} =\dfrac{\sqrt{1-\frac{1}{9}}}{\frac{1}{3}}= \dfrac{\sqrt{\frac{8}{9}}}{\frac{1}{3}} =\dfrac{\sqrt{8}} {3}:\dfrac{1}{3}=\sqrt{8}$