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

2 Answers

1/sqrt (1 + x^2 )11+x2

Explanation:

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

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

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

x^2=(sin(y)^2)/(cos(y)^2)x2=sin(y)2cos(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 !

Aug 3, 2018

Continuation of my first answer that was edited by another.

Explanation:

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,