Question

sin⁡(arcsin⁡2/3+arcsin 1/9) срочно помогите решение не из ФОТОМАТЧ

Answer 1

Ответ:

$sin(arccsin\dfrac{2}{3}+arcsin\dfrac{1}{9})=?\\\\\\\alpha =arcsin\dfrac{2}{3}\ \ \to \ \ \ sin\alpha =sin(arccsin\dfrac{2}{3})=\dfrac{2}{3}\\\\\beta =arcsin\dfrac{1}{9}\ \ \ \to\ \ \ sin\beta =sin(arcsin\dfrac{1}{9})=\dfrac{1}{9}\\\\\\cos^2\alpha =1-sin^2\alpha =1-\dfrac{4}{9}=\dfrac{5}{9}\ \ ,\ \ cos\alpha =\dfrac{\sqrt5}{3}\\\\\\cos^2\beta =1-sin^2\beta =1-\dfrac{1}{81}=\dfrac{80}{81}\ \ ,\ \ cos\beta =\dfrac{\sqrt{80}}{9}$

$sin(\alpha +\beta)=sin\alpha \cdot cos\beta +cos\alpha \cdot sin\beta =\dfrac{2}{3}\cdot \dfrac{\sqrt{80}}{9}+\dfrac{\sqrt5}{3}\cdot \dfrac{1}{9}=\\\\\\=\dfrac{2\sqrt{80}+\sqrt5}{27}=\dfrac{8\sqrt5+\sqrt5}{27}=\dfrac{9\sqrt5}{27}=\dfrac{\sqrt5}{3}\\\\\\sin(arccsin\dfrac{2}{3}+arcsin\dfrac{1}{9})=\dfrac{\sqrt5}{3}$