Question

Найдите значение трех основных тригонометрических функций а так же sin2 и cos2 .Помогите!!!
cos aльфа= √6/4 ; Πи/2<альфа<Πи

Answer 1

$\cos\alpha=\frac{\sqrt{6}}{4};\ \ \frac\pi2<\alpha<\pi;\\ \forall \alpha\in\left(\frac\pi2;\pi\right):\\ ctg\alpha<0;\ \ tg\alpha<0;\ \ \sin\alpha<0;\ \ \cos\alpha>0;\\ \sin\alpha=-\sqrt{1-\cos^2\alpha}=-\sqrt{1-\left(\frac{\sqrt6}{4}\right)^2}=-\sqrt{1-\frac{6}{16}}=\\ =-\sqrt{\frac{16}{16}-\frac6{16}}=-\sqrt{\frac{16-6}{16}}=-\sqrt{\frac{10}{16}}=-\frac{\sqrt{10}}{4};\\ tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{-\frac{\sqrt{10}}{4}}{\frac{\sqrt{6}}{4}}=-\sqrt{\frac{5}{3}};$
$ctg\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac1{tg\alpha}=\frac1{-\sqrt{\frac{5}{3}}}=-\sqrt{\frac35};\\ \sin2\alpha=2\cdot\sin\alpha\cdot\cos\alpha=2\cdot\left(\frac{-\sqrt{10}}{4}\right)\cdot\frac{\sqrt{6}}{4}=-\frac{2\cdot\sqrt{10}\cdot\sqrt{6}}{16}=-\frac{\sqrt{60}}{8}=\\ =-\frac{\sqrt{4\cdot15}}{8}=-\frac{2\sqrt{15}}{8}=-\frac{\sqrt{15}}{4}.\\ \cos2\alpha=2\cos^2\alpha-1=2\cdot\left(\frac{\sqrt6}{4}\right)^2-1=2\cdot\frac{6}{16}-1=\frac68-1=\\ =\frac68-\frac88=\frac{6-8}{8}=\frac{-2}{8}=-\frac14;$