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How do you simplify #cos(2 tan ^-1 x)#?

Answer 1

Use double angle formula to remove coefficient inside the cos, then rearrange standard trig definitions to make the trig function match the inverse trig function inside the bracket

Recall the double angle formula: #cos2theta=1-2sin^2theta#

Then #cos(2arctanx)=1-2sin^2arctanx#. NB I've written "arctan" here rather than "#tan^(-1)#" because the combination of exponents meaning powers and function inverses is potentially confusing.

So we now have a trig function of an inverse trig function. If we can express our #sin# in terms of #tan#, this will cancel right out.

By definition, #tantheta=(sintheta)/(costheta)=(sintheta)/sqrt(1-sin^2theta)#, so #tan^2theta(1-sin^2theta)=sin^2theta# #tan^2theta=sin^2theta(1+tan^2theta)# #sin^2theta=tan^2theta/(1+tan^2theta)#

By definition, #tanarctanx=x#, so #1-2sin^2arctanx# becomes #1-(2x^2)/(1+x^2)#. Putting this over a common denominator makes #(1-x^2)/(1+x^2)#.

So #cos(2arctanx)=(1-x^2)/(1+x^2)#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.