How do you simplify #cos(arctan(x)) #?

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Answer 1 · 2016-07-26T09:09:28

#1/sqrt (1 + x^2 )#

Let simplify #cos(Arctan(x))#
Let #y = arctan (x)#
#<=>x=tan(y)#

#x=sin(y)/cos(y)#

We need to have an expression for #cos(y)# only,

#x^2=(sin(y)^2)/(cos(y)^2)#

#x^2+1=cancel(sin(y)^2+cos(y)^2)^(=1)/cos(y)^2#

#1/(x^2+1)=cos(y)^2#

#1/sqrt(x^2+1)=cos(y)=cos(Arctan(x))#

\0/ Here's our answer !

Answer 2 · 2018-08-03T01:30:52

Continuation of my first answer that was edited by another.

Let a = arctan ( x ).

The principal value of #a in ( − π/2, π/2 )#.

Then #tan a = x# and

#0 <= cos a in [ 0, 1 )# ( wrongly marked as [ 0, - 1 ], in my

previous answer, two years ago).

Now, the given expression is

#cos a = 1/sqrt ( 1 + x^2 ), x in ( - pi/2, pi/2 )#.

It is important that #cos a >= 0#, for #a in Q_1# or #Q_4#.

If the piecewise-wholesome general inverse operator

#(tan)^( - 1 ) #is used,

#cos (tan)^(-1) x, = +-1/sqrt( 1 + x^2)#

the negative sign is chosen, when #x in Q_3#.

Example:

#cos (arctan 1 ) = 1/sqrt 2, arctan 1 = pi/4.#

#cos (tan)^(-1) 1 = cos ( kpi + pi/4), k = 0, +-1, +-2, +-, ...#

#in Q_1# or #Q_3, x = ...pi/4, 5/4pi, ...#

So, the value is #+-1/sqrt 2#..

My intention, in this approach, is to inform about nuances..

So, I am making a second answer,