Question

1.найти arccos(sin(-22п/5)) 2.выражение sin3a/sina-cos3a/cosa равно. 3.выражение cos(arctg1+arcsin12/13) равно.

Answer 1

$1)arccos(sin(-\frac{22\pi}{5}))=arccos(sin(-\frac{2\pi}{5}-4\pi))=$

$=arccos(sin(-\frac{2\pi}{5}))=arccos(cos(\frac{\pi}{2}-\frac{2\pi}{5}))=\frac{\pi}{10}$

$2)\frac{sin3a}{sina}-\frac{cos3a}{cosa}=\frac{sin3acosa-cos3asina}{sinacosa}=$

$=\frac{2sin(3a-a)}{sin2a}=\frac{2sin(2a)}{sin2a}=2$

$3)cos(arctg1+arcsin\frac{12}{13})=cos(\frac{\pi}{4}+arcsin\frac{12}{13})=$

$=\frac{\sqrt{2}}{2}cosarcsin\frac{12}{13}-\frac{\sqrt{2}}{2}sinarcsin\frac{12}{13}=$

$=\frac{\sqrt{2}}{2}(cosarccos(\frac{3\pi}{2}+\frac{12}{13})-\frac{12}{13})=$

$=\frac{\sqrt{2}}{2}\frac{3\pi}{2}=\frac{3\sqrt{2}\pi}{4}$