How do you simplify cos⁡(arctan⁡(x))?

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Melinda McCombs · Accepted Answer

Explanation:  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,  x2=sin⁡(y)2cos⁡(y)2  x2+1=sin⁡(y)2+cos⁡(y)2cos⁡(y)2  1x2+1=cos⁡(y)2  1x2+1=cos⁡(y)=cos⁡(arctan⁡(x))

Buck Henry · Accepted Answer

Explanation:  Let a=arctan⁡(x)  The principal value of a∈(−π2,π2)  Then tan⁡α=x and  0≤cos⁡α∈[0,1) (wrongly marked as [0,-1], in my previous answer, two years ago).  Now the given expression is  cos⁡α=11+x2,x∈(−π2,π2)  It is important that cos⁡α≥0, for α∈Q1 or Q4  If the piecewise-wholesome general inverse operator  (tan−1) is used  cos⁡(tan)−1x,=±11+x2  the negative sign is chosen, when x∈Q3  Example:  cos⁡(arctan⁡1)=12,arctan⁡1=π4  cos⁡(tan)−1=cos⁡(kπ+π4,k=0,±1,±2,±,…  ∈Q1 or Q3,x=…π4,54π,…  So, the value is ±12  My intention, in this approach, is to inform about nuances.

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

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