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

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

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: $\dfrac{1-8 \sqrt{3} }{15}$