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)#