anonymous
  • anonymous
cos(2tan-1(x))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
its arc-tan, the -1 is suppose to be a ^-1
anonymous
  • anonymous
must use \[cos(2\theta)=cos^2(\theta)-sin^2(\theta)\] so you need \[cos(tan^{-1}(x))\] and \[sin(tan^{-1}(x))\]
anonymous
  • anonymous
do you know how to find them?

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anonymous
  • anonymous
no
anonymous
  • anonymous
ok do this { draw a right triangle and label one angle as theta, theta =\[tan^{-1}(x)\] the angle whose tangent is x
anonymous
  • anonymous
since tangent is \[\frac{opp}{adj}\] label the opposite angle x and the adjacent angle as 1 so you have \[\frac{x}{1}=x\] as the tangent of your angle
anonymous
  • anonymous
by pythagoras the hypotenuse is\[\sqrt{x^2+1}\]
anonymous
  • anonymous
\[sin(tan^{-1}(x))=\frac{opp}{hyp}=\frac{x}{\sqrt{x^2+1}}\]
anonymous
  • anonymous
\[cos(tan^{-1}(x))=\frac{adj}{hyp}=\frac{1}{\sqrt{x^2+1}}\]
anonymous
  • anonymous
now use the "double angle" formula to find \[cos(2tan^{-1}(x))\]
anonymous
  • anonymous
it is the top line i wrote. calling \[tan^{-1}(x)=\theta\] it is \[cos^2(\theta)-sin^2(\theta)\]
anonymous
  • anonymous
you get \[\frac{1}{x^2+1}-\frac{x^2}{x^2+1}=\frac{1-x^2}{x^2+1}\]
anonymous
  • anonymous
the important part is being able to find \[sin(tan{-1}(x))\] or \[sin(tan{-1}(x))\]
anonymous
  • anonymous
i think i made a mistake on the last post where i said \[cos(tan^{-1}(x))\] was something else. it is \[\frac{1}{\sqrt{x^2+1}}\]

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