How do you find the exact value of $cos (arctan (\frac{5}{2}))$ $cos(arctan (5/2))$ ?

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Answer 1 · 2016-01-26T09:26:20

$cos (arctan (\frac{5}{2})) = \frac{2 \sqrt{29}}{29}$ $cos (arctan(5/2))=(2sqrt29)/29$

Let $A$ $A$ be an angle whose tangent $= \frac{5}{2}$ $=5/2$

Let $A = arctan (\frac{5}{2})$ $A=arctan(5/2)$

Then $tan A = \frac{5}{2}$ $tan A=5/2$

Imagine a right triangle with opposite side $a = 5$ $a=5$ and adjacent side $b = 2$ $b=2$ . Compute hypotenuse $c$ $c$

$c = \sqrt{a^{2} + b^{2}}$ $c=sqrt(a^2+b^2)$

$c = \sqrt{5^{2} + 2^{2}}$ $c=sqrt(5^2+2^2)$

$c = \sqrt{29}$ $c=sqrt29$

Then, the cosine function of $A$ $A$

$cos A = \frac{b}{c} = \frac{2}{\sqrt{29}} = \frac{2 \sqrt{29}}{29}$ $cos A=b/c=2/sqrt29=(2sqrt29)/29$

Have a nice day !!! From the Philippines...

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