[tex]\bf cos(x)=\cfrac{8}{17}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}
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\textit{using the pythagorean theorem}
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c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\implies \sqrt{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}
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c^2=a^2+b^2\implies\sqrt{c^2-b^2}=a\implies \sqrt{37^2-12^2}=a\implies 35=a[/tex]
[tex]\bf \\\\
-----------------------------\\\\
sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})\qquad thus
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sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
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sin(x+y)=\cfrac{15}{17}\cdot \cfrac{35}{37}+\cfrac{8}{17}\cdot \cfrac{12}{37}
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sin(x+y)=\cfrac{525}{629}+\cfrac{96}{629}\implies sin(x+y)=\cfrac{525+96}{629}
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sin(x+y)=\cfrac{621}{629}[/tex]
