How do you evaluate #cos[arc tan(-2/3)]#?

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Answer 1 · 2016-08-14T02:35:32

#3/sqrt 13#

Let #a = arc tan (-2/3) in Q4, by the convention of choosing the

principal value, when the tangent is negative. In Q4, cosine is

positive,

Now, the given expression is

#cos a =3/sqrt(2^2+3^2)=3/sqrt 13#.

Had we chosen #a in Q4#, from the general value ( wherein cosine is

negative ), the answer would become #-3/sqrt 13#.