Question

Решите пожалуйста задание на фотографии! !!!!!!!! очень срочно

Answer 1

1)
$sin a=-\sqrt{1-cos^2 a}=-\sqrt{1-(\frac{\sqrt{5}}{5})^2}=\\\\-\frac{2}{\sqrt{5}}$
$tg a=\frac{sin a}{cos a}=\\\\-\frac{2}{\sqrt{5}}:\frac{\sqrt{5}}{5}=-10$

2)
$cos^2 a+sin^2 a=1; 1+ctg^2 a=\frac{1}{sin^2 a}$
$sin a=-\frac{1}{\sqrt{1+ctg^2 a}}=\\\\-\frac{1}{\sqrt{1+(-\frac{\sqrt{21}}{2})^2}}=-0.4$

3)
$sin a=-\sqrt{1-cos^2 a}=-\sqrt{1-(-\frac{1}{\sqrt{17}})^2}=\\\\ -\frac{4}{\sqrt{17}}$
$tg a=\frac{sin a}{cos a}=\frac{-4}{\sqrt{17}}:\frac{-1}{\sqrt{17}}=4$

4)
$cos a=-\sqrt{1-sin^2 a}=-\sqrt{1-(\frac{\sqrt{2}}{2})^2}=-\frac{\sqrt{2}}{2}$
$\frac{1}{2}+\sqrt{2}cos a=\frac{1}{2}+\sqrt{2}*\frac{-\sqrt{2}}{2}=-\frac{1}{2}$

5)
 $sin a=-\sqrt{1-cos^2 a}=-\sqrt{1-(\frac{\sqrt{2}}{2})^2}=-\frac{\sqrt{2}}{2}$
$\frac{\sqrt{2}}{10}sin a+2=\frac{\sqrt{2}}{10}*(-\frac{\sqrt{2}}{2})+2=1.9$

6)
 $\frac{\sqrt{13}}{2}cos(\frac{\pi}{2}-a)=\frac{\sqrt{13}}{2}sin a=\\\\-\frac{\sqrt{13}}{2}*\sqrt{1-cos^2 a}=\\\\-\frac{\sqrt{13}}{2}*(\sqrt{1-(\sqrt{\frac{5}{13}})^2})=\\\\-\frac{\sqrt{13}}{2}*\frac{12}{\sqrt{13}}=-6$

7)
 $1-\sqrt{\frac{14}{3}}sin(a+\pi)=1-\sqrt{\frac{14}{3}}*sin a=\\\\1-\sqrt{\frac{14}{3}}*\sqrt{1-cos^2 a}=1-\sqrt{\frac{14}{3}}*(\sqrt{1-(\frac{1}{\sqrt{7}})^2})=\\\\1-\frac{\sqrt{14}}{\sqrt{3}}*\frac{\sqrt{6}}{\sqrt{7}}=1-2=-1$

8)
$\sqrt{7}cos(\pi-a)-\frac{1}{2}=\sqrt{7}*(-cos a)-\frac{1}{2}=\\\\ \sqrt{7}*(-(-\sqrt{1-cos^2 a})-\frac{1}{2}=\\\\ \sqrt{7}*\sqrt{1-(-\frac{\sqrt{3}}{\sqrt{7}})^2}-\frac{1}{2}=\\\\ \sqrt{7}*\frac{2}{\sqrt{7}}-0.5=2-0.5=1.5$

Answer 2

$tg a=\frac{sin a}{cos a}=\frac{-\sqrt{1-cos^2 a}}{cos a}= \frac{- \sqrt{1- \frac{1}{5} } }{\frac{\sqrt{5}}{5}}=-2$

$1+ctg^2 a=\frac{1}{sin^2 a} \\\ sin a=-\frac{1}{\sqrt{1+ctg^2 a}}=-\frac{1}{\sqrt{1+(-\frac{\sqrt{21}}{2})^2}}=-0.4$

$sin a=-\sqrt{1-cos^2 a}=-\sqrt{1-(-\frac{1}{\sqrt{17}})^2}=-\frac{4}{\sqrt{17}} \\\ tg a=\frac{sin a}{cos a}= \frac{\frac{-4}{\sqrt{17}}}{\frac{-1}{\sqrt{17}}}=4$

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

$\frac{\sqrt{2}}{10}sin a+2\frac{\sqrt{2}}{10}\cdot(-\sqrt{1-cos^2 a})+2 =\frac{\sqrt{2}}{10}\cdot(-\sqrt{1-(\frac{\sqrt{2}}{2})^2}})+2=1.9$

$\sqrt{ \frac{13}{2}}cos(\frac{\pi}{2}-a)= \sqrt{ \frac{13}{2}}sin a=- \sqrt{ \frac{13}{2}}\sqrt{1-cos^2 a}= \\\ =- \sqrt{ \frac{13}{2}}\sqrt{1-(\sqrt{\frac{5}{13}})^2}=- \sqrt{ \frac{13}{2}}\cdot\sqrt{ \frac{8}{13}}=-2$

$1-\sqrt{\frac{14}{3}}sin(a+\pi)=1-\sqrt{\frac{14}{3}}sin a=1-\sqrt{\frac{14}{3}}\sqrt{1-cos^2 a}= \\\ =1-\sqrt{\frac{14}{3}}(\sqrt{1-(\frac{1}{\sqrt{7}})^2})=1-\sqrt{\frac{14}{3}}\cdot\frac{\sqrt{6}}{\sqrt{7}}=1-2=-1$

$\sqrt{7}cos(\pi-a)-\frac{1}{2}=\sqrt{7}(-cos a)-\frac{1}{2}=\sqrt{7}(-(-\sqrt{1-cos^2 a})-\frac{1}{2}= \\\ =\sqrt{7}\sqrt{1-(-\frac{\sqrt{3}}{\sqrt{7}})^2}-\frac{1}{2}=\sqrt{7}\frac{2}{\sqrt{7}}-\frac{1}{2}=2-\frac{1}{2}=\frac{3}{2}$