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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)

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  1. jim_thompson5910
    • one year ago
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    Draw a right triangle |dw:1436829693684:dw|

  2. jim_thompson5910
    • one year ago
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    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?

  3. anonymous
    • one year ago
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    yup :)

  4. jim_thompson5910
    • one year ago
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    what is the length of the hypotenuse?

  5. anonymous
    • one year ago
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    r=1^2+x^2 so x^2?

  6. jim_thompson5910
    • one year ago
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    close, but the square root of 1^2+x^2 or the square root of 1+x^2 |dw:1436830005407:dw|

  7. anonymous
    • one year ago
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    oh okay

  8. jim_thompson5910
    • one year ago
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    with that triangle, you can find sin(theta) remember I made theta be equal to the inverse tangent of x

  9. anonymous
    • one year ago
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    cos=y/r 1/sqrtx^2+1 1/x+1

  10. anonymous
    • one year ago
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    hmm I think I did something wrong

  11. jim_thompson5910
    • one year ago
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    you want sine, so sine = opposite/hypotenuse \[\Large \sin(\theta) = \frac{x}{\sqrt{x^2+1}}\] \[\Large \sin\left(\tan^{-1}(x)\right) = \frac{x}{\sqrt{x^2+1}}\]

  12. jim_thompson5910
    • one year ago
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    |dw:1436830701065:dw|

  13. jim_thompson5910
    • one year ago
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    btw, \[\Large \sqrt{x^2+1} \ne x+1\] \[\Large \sqrt{x^2+y^2} \ne x+y\]

  14. jim_thompson5910
    • one year ago
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    and \[\Large \sqrt{A+B} \ne \sqrt{A} + \sqrt{B}\]

  15. anonymous
    • one year ago
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    oh sorry I didn't know that :o

  16. anonymous
    • one year ago
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    how would I simplify it?

  17. jim_thompson5910
    • one year ago
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    you can rationalize the denominator, but you can't really do anything else to simplify

  18. anonymous
    • one year ago
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    I would rationalize it by multiplying both sides by sqrtx^2+1 right?

  19. jim_thompson5910
    • one year ago
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    top and bottom by sqrt(x^2+1)

  20. anonymous
    • one year ago
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    ohh okay I understand the problem now :)

  21. jim_thompson5910
    • one year ago
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    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

  22. anonymous
    • one year ago
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    I see

  23. anonymous
    • one year ago
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    thank for all your help! I get it now :D sorry it took so much time

  24. jim_thompson5910
    • one year ago
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    that's ok and you're welcome

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