hey guys, need help is solving cos(2tan^-1x)

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hey guys, need help is solving cos(2tan^-1x)

Mathematics
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what do you have to do with that function...
simplify probably :)
yea fair call, probs a good idea to apply what arctanx means

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\[\large\rm \cos(2\color{orangered}{\arctan x})\] If \(\large\rm arctan x=\theta\) then \(\large\rm \tan\theta=x\) This arctangent is just some angle.\[\large\rm \cos(2\color{orangered}{\arctan x})=\cos(2\color{orangered}{\theta})\]So we need to apply our Double Angle Formula for Cosine. It shows up in three different forms, any of them will do.\[\large\rm \cos(2\theta)=2\cos^2\theta-1\]
We should draw a triangle to show what is going on with this tangent function.
\[\large\rm \tan \theta=x\qquad\to\qquad \tan\theta=\frac{x}{1}=\frac{opposite}{adjacent}\]
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Then use Pythagorean Theorem to find the missing hypotenuse.
ye lel i'll delete mine
And use that information to finish it off
probs a good idea to realise \[\theta=\tan ^{-1}x\] and you therefore wants to find \[\cos(2\theta)\]
then use trig identities
all g?
thanks guys.

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