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anonymous
 one year ago
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)
anonymous
 one year ago
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)

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Draw a right triangle dw:1436829693684:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1If 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?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1what is the length of the hypotenuse?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1close, but the square root of 1^2+x^2 or the square root of 1+x^2 dw:1436830005407:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1with that triangle, you can find sin(theta) remember I made theta be equal to the inverse tangent of x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos=y/r 1/sqrtx^2+1 1/x+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm I think I did something wrong

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436830701065:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1btw, \[\Large \sqrt{x^2+1} \ne x+1\] \[\Large \sqrt{x^2+y^2} \ne x+y\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1and \[\Large \sqrt{A+B} \ne \sqrt{A} + \sqrt{B}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh sorry I didn't know that :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would I simplify it?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you can rationalize the denominator, but you can't really do anything else to simplify

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would rationalize it by multiplying both sides by sqrtx^2+1 right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1top and bottom by sqrt(x^2+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay I understand the problem now :)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Rationalizing 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
 one year ago
Best ResponseYou've already chosen the best response.0thank for all your help! I get it now :D sorry it took so much time

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1that's ok and you're welcome
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