What does $cos(arctan((3pi)/4))$ equal?

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Answer 1 · 2018-03-11T11:58:51

$cos(arctan((3pi)/4)) = 4/sqrt(9pi^2+16)$

This is a rather curious question, since $(3pi)/4$ looks like an angle and not the result of applying $tan$ to an angle.

Nevertheless $(3pi)/4$ is a value taken by $tan$ somewhere in Q1.

Consider a right angled triangle with angle $theta$ and sides $"opposite" = 3pi$ , $"adjacent" = 4$ and $"hypotenuse" = sqrt(9pi^2+16)$ .

Then:

$tan theta = "opposite"/"adjacent" = (3pi)/4$

$cos theta = "adjacent"/"hypotenuse" = 4/sqrt(9pi^2+16)$

Then:

$cos(arctan((3pi)/4)) = cos theta = 4/sqrt(9pi^2+16)$

What does cos(arctan((3pi)/4)) cos(arctan((3pi)/4)) equal?