Question

Вычислите + решение
Sin(arcsin 1/3 - arccos 1/5)

Answer 1

$sin(arcsin dfrac{1}{3}-arccos dfrac{1}{5})= \ = sin(arcsin dfrac{1}{3})cdot cos(arccos dfrac{1}{5})-cos(arcsin dfrac{1}{3}) cdot sin(arccos dfrac{1}{5})= \ = dfrac{1}{3}cdot dfrac{1}{5} -cos(arcsin dfrac{1}{3}) cdot sin(arccos dfrac{1}{5})= \ = dfrac{1}{15}-cos(arcsin dfrac{1}{3}) cdot sin(arccos dfrac{1}{5})$

Пусть
$arcsin dfrac{1}{3}= alpha \ arccos dfrac{1}{5}= beta$
тогда
$sin alpha = dfrac{1}{3} \ cos beta = dfrac{1}{5}$
и
$cos alpha = sqrt{1-( dfrac{1}{3})^2 }= dfrac{2 sqrt{2} }{3} \ sin beta = sqrt{1- (dfrac{1}{5})^2 } = dfrac{2 sqrt{6} }{5}$
значит
$cos(arcsin dfrac{1}{3}) cdot sin(arccos dfrac{1}{5})= dfrac{2 sqrt{2} }{3}cdot dfrac{2 sqrt{6} }{5}= dfrac{8 sqrt{3} }{15}$
и, наконец
$dfrac{1}{15}-cos(arcsin dfrac{1}{3}) cdot sin(arccos dfrac{1}{5})= dfrac{1}{15}- dfrac{8 sqrt{3} }{15}= dfrac{1-8 sqrt{3} }{15}$

Ответ: $dfrac{1-8 sqrt{3} }{15}$