Вычеслить 1) cos (arcsin 2/9) 2)sin (arccos 3/4) 3),sin (arctg 3) 4)cos (arcctg (-2)) 5)...

Question 1

Вычеслить
1) cos (arcsin 2/9)
2)sin (arccos 3/4)
3),sin (arctg 3)
4)cos (arcctg (-2))
5) tg ( arcsin 1/5)
6)
ctg (arctg 6)

Question 2

Решите задачу:

$1)arcsin \frac{2}{9}=\alpha \to sin \alpha =sin(arcsin \frac{2}{9} )=\frac{2}{9}\\cos(arcsin\frac{2}{9})=cos \alpha =\sqrt{1-sin^2 \alpha }=\sqrt{1-\frac{4}{81}}=\frac{\sqrt{77}}{9}\\2)arccoc\frac{3}{4}= \alpha \to cos \alpha =\frac{3}{4}\\sin(arccos\frac{3}{4})=sin \alpha =\sqrt{1-cos^2 \alpha }=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\\3)arctg3= \alpha \to tg \alpha =3\\1+tg^2 \alpha =\frac{1}{cos^2 \alpha } \to cos^2 \alpha =\frac{1}{1+tg^2 \alpha }=\frac{1}{1+9}=\frac{1}{10}$
$sin \alpha =\sqrt{1-cos^2 \alpha }=\sqrt{1-\frac{1}{10}}=\frac{3}{\sqrt{10}}=sin(arctg3)\\4)cos(arcctg(-2))=cos \alpha , ctg \alpha =-2,\\1+ctg^2 \alpha =\frac{1}{sin^2 \alpha }\to sin^2 \alpha =\frac{1}{1+ctg^2 \alpha }=\frac{1}{1+4}=\frac{1}{5}\\cos \alpha =\sqrt{1-sin^2 \alpha} =\sqrt{1-\frac{1}{5}}=\frac{2}{\sqrt{5}}=cos(arcctg(-2))\\5)tg(arcsin\frac{1}{5})=tg \alpha ,\to sin \alpha =\frac{1}{5}\\cos \alpha =\sqrt{1-\frac{1}{25}}=\frac{\sqrt{24}}{5}\\tg \alpha =\frac{sin \alpha }{cos \alpha }$
$tg \alpha =\frac{1}{5}:\frac{\sqrt{24}}{5}=\frac{1}{\sqrt{24}}=tg(arcsin\frac{1}{5})\\6)ctg(arctg6)=ctg \alpha , \alpha =arctg6,\to tg \alpha =6\\ctg \alpha =\frac{1}{tg \alpha }=\frac{1}{6}$