How do you find the exact value of $cos [arctan (- \frac{5}{12})]$ $cos[arctan(-5/12)]$ ?

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How do you find the exact value of $cos [arctan (- \frac{5}{12})]$ $cos[arctan(-5/12)]$ ?

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Answer 1 · 2015-05-15T21:50:35

Notice that 5 and 12 are two sides of a right angled triangle whose hypotenuse is 13, since $5^{2} + 12^{2} = 25 + 144 = 169 = 13^{2}$ $5^2 + 12^2 = 25 + 144 = 169 = 13^2$

So if $θ$ $theta$ is the smallest angle in the $5$ $5$ , $12$ $12$ , $13$ $13$ triangle then

$sin θ = \frac{5}{13}$ $sin theta = 5/13$ , $cos θ = \frac{12}{13}$ $cos theta = 12/13$ and $tan θ = \frac{5}{12}$ $tan theta = 5/12$ .

Then $tan (- θ) = - tan (θ) = - \frac{5}{12}$ $tan (-theta) = -tan(theta) = -5/12$

So we are looking for $cos (- θ) = cos (θ) = \frac{12}{13}$ $cos (-theta) = cos(theta) = 12/13$

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How do you find the exact value of $cos [arctan (- \frac{5}{12})]$ $cos[arctan(-5/12)]$ ?

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