[tex]\bf cos\left[tan^{-1}\left(\frac{12}{5} \right)+ tan^{-1}\left(\frac{-8}{15} \right) \right]\\
\left. \qquad \qquad \quad \right.\uparrow \qquad \qquad \qquad \uparrow \\
\left. \qquad \qquad \quad \right.\alpha \qquad \qquad \qquad \beta
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\textit{that simply means }tan(\alpha)=\cfrac{12}{5}\qquad and\qquad tan(\beta)=\cfrac{-8}{5}
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\textit{so, we're really looking for }cos(\alpha+\beta)[/tex]
now.. hmmm -8/15 is rather ambiguous, since the negative sign is in front of the rational, and either 8 or 15 can be negative, now, we happen to choose the 8 to get the minus, but it could have been 8/-15
ok, well hmm so, the issue boils down to
[tex]\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad thus
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tan(\alpha)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}
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\textit{so, what is the hypotenuse "c"?}\\
\textit{ well, let's use the pythagorean theorem}
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c=\sqrt{a^2+b^2}\implies c=\sqrt{25+144}\implies c=\sqrt{169}\implies \boxed{c=13}\\\\
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\textit{this simply means }\boxed{cos(\alpha)=\cfrac{5}{13}\qquad \qquad sin(\alpha)=\cfrac{12}{13}
}[/tex]
now, let's take a peek at the second angle, angle β
[tex]\bf tan(\beta)=\cfrac{-8}{15}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}
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\textit{again, let's find "c", or the hypotenuse}
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c=\sqrt{15^2+(-8)^2}\implies c=\sqrt{289}\implies \boxed{c=17}\\\\
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thus\qquad \boxed{cos(\beta)=\cfrac{15}{17}\qquad \qquad sin(\beta)=\cfrac{-8}{17}}[/tex]
now, with that in mind, let's use the angle sum identity for cosine
[tex]\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})\\\\
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cos({{ \alpha}} + {{ \beta}})= \left( \cfrac{5}{13} \right)\left( \cfrac{15}{17} \right)-\left( \cfrac{12}{13} \right)\left( \cfrac{-8}{17} \right)
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cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}-\cfrac{-96}{221}\implies cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}+\cfrac{96}{221}
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\boxed{cos({{ \alpha}} + {{ \beta}})=\cfrac{171}{221}}[/tex]
