How do you simplify the expression $cos (arctan (\frac{x}{5}))$ $cos(arctan(x/5))$ ?

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How do you simplify the expression $cos (arctan (\frac{x}{5}))$ $cos(arctan(x/5))$ ?

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Answer 1 · 2016-11-10T16:12:45

$cos (arctan (\frac{x}{5})) = \pm \sqrt[2]{\frac{1}{1 + {(\frac{x}{5})}^{2}}}$ $cos(arctan(x/5))=+-root2(1/(1+(x/5)^2)$

Explanation:

From the fundamental identity of trigonometry ${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$
we can deduce by dividing both sides for ${cos}^{θ}$ $cos^theta$ and imposing the existence condition $θ \neq \frac{π}{2} + k π$ $theta ne pi/2+kpi$

$1 + {tan}^{2} θ = \frac{1}{{cos}^{2} θ}$ $1+tan^2theta=1/cos^2theta$ that can be rewritten as

${cos}^{2} θ = \frac{1}{1 + {tan}^{2} θ}$ $cos^2theta=1/(1+tan^2theta)$ or

$cos θ = \pm \sqrt[2]{\frac{1}{1 + {tan}^{2} θ}}$ $costheta=+-root2(1/(1+tan^2theta)$
if $θ$ $theta$ is $arctan (\frac{x}{5})$ $arctan (x/5)$ then
$cos (arctan (\frac{x}{5})) = \pm \sqrt[2]{\frac{1}{1 + {tan}^{2} (arctan (\frac{x}{5}))}}$ $cos(arctan(x/5))=+-root2(1/(1+tan^2(arctan(x/5))$
but $tan (arctan (\frac{x}{5})) = \frac{x}{5}$ $tan(arctan(x/5))=x/5$
in the end we can finally write
$cos (arctan (\frac{x}{5})) = \pm \sqrt[2]{\frac{1}{1 + {(\frac{x}{5})}^{2}}}$ $cos(arctan(x/5))=+-root2(1/(1+(x/5)^2)$

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How do you simplify the expression $cos (arctan (\frac{x}{5}))$ $cos(arctan(x/5))$ ?

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How do you simplify the expression cos(arctan(x5))cos(arctan(x/5)) ?

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How do you simplify the expression $cos (arctan (\frac{x}{5}))$ $cos(arctan(x/5))$ ?