Question

1.Найти arccos(sin(-22П/5))

2.Выражение sin3a/sina-cos3a/cosa равно.

3.Выражение cos(arctg1+arcsin12/13) равно.

Answer 1

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

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

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

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

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

$=frac{sqrt{2}}{2}cosarcsinfrac{12}{13}-frac{sqrt{2}}{2}sinarcsinfrac{12}{13}=$

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

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