Question

How would I go about solving this?
`sin(x+y)` if cosx=8/17 and siny=12/37

Answer 1

`\bf cos(x)=\cfrac{8}{17}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c} \\\\\\ \textit{using the pythagorean theorem} \\\\\\ c^2=a^2+b^2\implies √(c^2-a^2)=b\implies √(17^2-8^2)=b\implies 15=b\\\\ -----------------------------\\\\ sin(y)=\cfrac{12}{37}\cfrac{\leftarrow opposite=b}{\leftarrow hypotenuse=c}\\\\\\ \textit{using the pythagorean theorem} \\\\\\ c^2=a^2+b^2\implies√(c^2-b^2)=a\implies √(37^2-12^2)=a\implies 35=a`


`\bf \\\\ -----------------------------\\\\ sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})\qquad thus \\\\\\ sin(x+y)=sin(x)cos(y)+cos(x)sin(y) \\\\\\ sin(x+y)=\cfrac{15}{17}\cdot \cfrac{35}{37}+\cfrac{8}{17}\cdot \cfrac{12}{37} \\\\\\ sin(x+y)=\cfrac{525}{629}+\cfrac{96}{629}\implies sin(x+y)=\cfrac{525+96}{629} \\\\\\ sin(x+y)=\cfrac{621}{629}`