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Math Help - Verifying Trig Identities

  1. #1
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    Verifying Trig Identities

    I understand the concept and know the identities, however our teacher offered us more difficult bonus problems and this one threw me for a loop.

    The problem is

    the square root of (1-sin(theta))/(1+sin(theta)) = 1/(sec(thetha) - tan(theta))

    My question is... can you square both sides? I tried that and worked the right side until I got (1-sin^2(theta))/(1+sin^2(theta)) ... which, if I took the square root of, I would get (1-sin(theta))/(1+sin(theta)) however I can't just take the square root of one side.

    I really don't know where to go with this problem..

    Any help is appreciated!

    Also, sorry if it makes no sense when I wrote it out.. if you would like me to write it out as an image so it makes more sense, I can. Thanks!
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  2. #2
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    Quote Originally Posted by mgannon93 View Post
    I understand the concept and know the identities, however our teacher offered us more difficult bonus problems and this one threw me for a loop.

    The problem is

    the square root of (1-sin(theta))/(1+sin(theta)) = 1/(sec(thetha) - tan(theta))

    My question is... can you square both sides? I tried that and worked the right side until I got (1-sin^2(theta))/(1+sin^2(theta)) ... which, if I took the square root of, I would get (1-sin(theta))/(1+sin(theta)) however I can't just take the square root of one side.

    I really don't know where to go with this problem..

    Any help is appreciated!

    Also, sorry if it makes no sense when I wrote it out.. if you would like me to write it out as an image so it makes more sense, I can. Thanks!
    Please, check if it is 1/ there.
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  3. #3
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    Here it is typed out. (Hopefully this works, I'm posting it as an image).
    Does this make more sense?

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  4. #4
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    Quote Originally Posted by mgannon93 View Post
    Here it is typed out. (Hopefully this works, I'm posting it as an image).
    Does this make more sense?

    AHHHH1!!! It's +tan x!!!!
    OK.
    multiply the numerator and the denominator on the left side by (1-sinx)
    ROOT of fraction
    (1-sinx)(1-sinx)/ 1/
    (1+sinx)(1-sinx) = 1/ sinx/
    cosx + cosx

    ROOT
    numerator(1-sin)^2 / = 1/
    denominator(1^2-sin^2x) (1+sinx)/
    cosx

    by sin^2x+cos^2x=1(Pythagorean Identity) denom is cos^2x
    taking root we have on the left
    (1-sinx)/cosx

    On the right: 1 divide by fraction (1+sinx)/cosx is 1multiply by reciprocal cosx/(1+sinx). So it's cosx/(1+sinx).
    We've got a proportion:
    (1-sinx) cosx
    cosx = (1+sinx)

    Cross multiply we get (1-sinx)(1+sinx)=cos^2x, open parenthesses and we get 1-sin^2x=cos^2x(Pythogorean Identity)
    Last edited by tdschool; November 5th 2008 at 06:46 PM.
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  5. #5
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    Last edited by tdschool; November 18th 2008 at 03:48 PM.
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  6. #6
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    Hello, mgannon93!

    \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} \;=\;\frac{1}{\sec\theta - \tan\theta}

    \sqrt{\frac{1-\sin\theta}{1+\sin\theta}\cdot{\color{blue}\frac{1  +\sin\theta}{1+\sin\theta}}} \;=\;\sqrt{\frac{1-\sin^2\!\theta}{(1+\sin\theta)^2}} \;=\;\sqrt{\frac{\cos^2\!\theta}{(1+\sin\theta)^2}  } . = \;\frac{\cos\theta}{1 + \sin\theta}


    Divide top and bottom by \cos\theta\!:\quad \frac{\dfrac{\cos\theta}{\cos\theta}}{\dfrac{1}{\c  os\theta} + \dfrac{\sin\theta}{\cos\theta}} \;=\;\frac{1}{\sec\theta + \tan\theta}<br /> <br />

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  7. #7
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    Thumbs up

    Thank you both so much!

    I get it now.

    Thanks!
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