Question

sin(альфа + бета) и cos (альфа - бета), если sina = 8/17, cosB =4/5, 0<a< П/2, 0<B<П/2
помогите пожалуйста ​

Answer 1

Объяснение:

$1)\ sin\alpha =\frac{8}{17} \ \ \ \ 0\leq \alpha \leq \frac{\pi }{2}\\ sin^2\alpha +cos^2\alpha =1\\cos^2\alpha =1-sin^2\alpha =1-(\frac{8}{17})^2=1-\frac{64}{289}=\frac{289-64}{289}=\frac{225}{289}.\\ cos\alpha =б\sqrt{\frac{225}{289} }=б\frac{15}{17} .\ \ \ \ 0\leq \alpha \leq \frac{\pi }{2}\ \ \ \ \ \Rightarrow\\ cos\alpha =\frac{15}{17} .$

$2)\ cos\beta =\frac{4}{5}\ \ \ \ 0\leq \beta \leq \frac{\pi }{2} \\sin^2\beta +cos^2\beta =1\\sin^2\beta =1-cos^2\beta =1-(\frac{4}{5})^2=1-\frac{16}{25}=\frac{25-16}{25}=\frac{9}{25} .\\ sin\beta =б\sqrt{\frac{9}{25} } = б\frac{3}{5} \ \ \ \ 0\leq \beta \leq \frac{\pi }{2}\ \ \ \ \Rightarrow\\ sin\beta =\frac{3}{5}.$

$3)\ sin(\alpha +\beta )=sin\alpha *cos\beta +cos\alpha *sin\beta =\frac{8}{17}*\frac{4}{5}+\frac{15}{17} *\frac{3}{5}=\frac{32}{85} +\frac{45}{85}=\frac{77}{85} .$

$4)\ cos(\alpha -\beta )=cos\alpha *cos\beta +sin\alpha *sin\beta =\frac{15}{17}*\frac{4}{5}+\frac{8}{17} *\frac{3}{5}=\frac{60}{85}+\frac{24}{85} =\frac{84}{85} .$

Ответ: sin(α+β)=77/85,  cos(α-β)=84/85.