anonymous
  • anonymous
what are the steps to solve this problem? Write each trigonometric expression as an algebraic expression in x 1)sin(tan^-1x) 2)tan(cos^-1x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
Draw a right triangle |dw:1436829693684:dw|
jim_thompson5910
  • jim_thompson5910
If we let \(\Large \theta = \tan^{-1}(x)\) then \(\Large \tan(\theta) = x = \frac{x}{1}\) because tangent = opposite/adjacent, we can add these labels |dw:1436829773980:dw| making sense so far?
anonymous
  • anonymous
yup :)

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jim_thompson5910
  • jim_thompson5910
what is the length of the hypotenuse?
anonymous
  • anonymous
r=1^2+x^2 so x^2?
jim_thompson5910
  • jim_thompson5910
close, but the square root of 1^2+x^2 or the square root of 1+x^2 |dw:1436830005407:dw|
anonymous
  • anonymous
oh okay
jim_thompson5910
  • jim_thompson5910
with that triangle, you can find sin(theta) remember I made theta be equal to the inverse tangent of x
anonymous
  • anonymous
cos=y/r 1/sqrtx^2+1 1/x+1
anonymous
  • anonymous
hmm I think I did something wrong
jim_thompson5910
  • jim_thompson5910
you want sine, so sine = opposite/hypotenuse \[\Large \sin(\theta) = \frac{x}{\sqrt{x^2+1}}\] \[\Large \sin\left(\tan^{-1}(x)\right) = \frac{x}{\sqrt{x^2+1}}\]
jim_thompson5910
  • jim_thompson5910
|dw:1436830701065:dw|
jim_thompson5910
  • jim_thompson5910
btw, \[\Large \sqrt{x^2+1} \ne x+1\] \[\Large \sqrt{x^2+y^2} \ne x+y\]
jim_thompson5910
  • jim_thompson5910
and \[\Large \sqrt{A+B} \ne \sqrt{A} + \sqrt{B}\]
anonymous
  • anonymous
oh sorry I didn't know that :o
anonymous
  • anonymous
how would I simplify it?
jim_thompson5910
  • jim_thompson5910
you can rationalize the denominator, but you can't really do anything else to simplify
anonymous
  • anonymous
I would rationalize it by multiplying both sides by sqrtx^2+1 right?
jim_thompson5910
  • jim_thompson5910
top and bottom by sqrt(x^2+1)
anonymous
  • anonymous
ohh okay I understand the problem now :)
jim_thompson5910
  • jim_thompson5910
Rationalizing the denominator gives... \[\Large \sin\left(\tan^{-1}(x)\right) = \frac{x}{\sqrt{x^2+1}}\] \[\Large \sin\left(\tan^{-1}(x)\right) = \frac{x{\color{red}{*\sqrt{x^2+1}}}}{\sqrt{x^2+1}{\color{red}{*\sqrt{x^2+1}}}}\] \[\Large \sin\left(\tan^{-1}(x)\right) = \frac{x\sqrt{x^2+1}}{(\sqrt{x^2+1})^2}\] \[\Large \sin\left(\tan^{-1}(x)\right) = \frac{x\sqrt{x^2+1}}{x^2+1}\] to me, that's not really simplified. If anything, it got more complicated. But some books require you to rationalize the denominator
anonymous
  • anonymous
I see
anonymous
  • anonymous
thank for all your help! I get it now :D sorry it took so much time
jim_thompson5910
  • jim_thompson5910
that's ok and you're welcome

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