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Thread: proving expressions.

  1. #1
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    proving expressions.

    Given that $\displaystyle a = 4 sec x , b = cosx , c = cotx $

    a) express b in terms of a
    b) show that $\displaystyle c^{2} = \frac{16}{a^{2} - 16} $
    a) $\displaystyle b = cosx $
    $\displaystyle a = \frac{4}{cosx} $
    $\displaystyle a = \frac{4}{b} $
    $\displaystyle b = \frac{4}{a} $

    having trouble with part 'b';
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  2. #2
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    Quote Originally Posted by Tweety View Post
    a) $\displaystyle b = cosx $
    $\displaystyle a = \frac{4}{cosx} $
    $\displaystyle a = \frac{4}{b} $
    $\displaystyle b = \frac{4}{a} $

    having trouble with part 'b';
    $\displaystyle \frac{16}{(4\sec{x})^2 - 16}$

    $\displaystyle \frac{16}{16\sec^2{x} - 16}$

    $\displaystyle \frac{1}{\sec^2{x} - 1}$

    $\displaystyle \frac{1}{\tan^2{x}}$

    $\displaystyle \cot^2{x} = c^2$
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  3. #3
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    Quote Originally Posted by skeeter View Post
    $\displaystyle \frac{16}{(4\sec{x})^2 - 16}$

    $\displaystyle \frac{16}{16\sec^2{x} - 16}$

    $\displaystyle \frac{1}{\sec^2{x} - 1}$

    $\displaystyle \frac{1}{\tan^2{x}}$

    $\displaystyle \cot^2{x} = c^2$
    I dont really understand what you did, aren't you meant to prove that $\displaystyle c^{2} = \frac{16}{a^{2}-16} $? Why are you doing $\displaystyle c^{2} = cot^2{x} $?

    This is what I have done but it does not fall together!

    $\displaystyle c = cotx , c^{2} = cot^{2}x $

    $\displaystyle b = \frac{4}{a} , b^{2} = \frac{16}{a^{2}} $

    $\displaystyle cos^{2}x = \frac{16}{a^{2}} $

    $\displaystyle 1-sin^{2}x = \frac{16}{a^{2}} $

    $\displaystyle a^{2}(1-sin^{2}x) = 16 $

    $\displaystyle a^{2} = \frac{16}{1-sin^{2}x} $

    $\displaystyle a^{2} = \frac{16}{1} - \frac{16}{sin^{2}x} $

    $\displaystyle a^{2} = 16 - 16cosec^2x $

    $\displaystyle a^2 = 16 - 16(1+cot^{2}x) $

    $\displaystyle a^{2}= 16- 16(1+c^{2}) $

    $\displaystyle a^{2} -16 =-16( 1+c^{2}) $


    is this right so far? But I am stuck here,
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  4. #4
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    Quote Originally Posted by Tweety View Post
    I dont really understand what you did, aren't you meant to prove that $\displaystyle c^{2} = \frac{16}{a^{2}-16} $? Why are you doing $\displaystyle c^{2} = cot^2{x} $?

    This is what I have done but it does not fall together!

    $\displaystyle c = cotx , c^{2} = cot^{2}x $

    $\displaystyle b = \frac{4}{a} , b^{2} = \frac{16}{a^{2}} $

    $\displaystyle cos^{2}x = \frac{16}{a^{2}} $

    $\displaystyle 1-sin^{2}x = \frac{16}{a^{2}} $

    $\displaystyle a^{2}(1-sin^{2}x) = 16 $

    $\displaystyle a^{2} = \frac{16}{1-sin^{2}x} $

    $\displaystyle a^{2} = \frac{16}{1} - \frac{16}{sin^{2}x} $ this step is incorrect ... you cannot "split" the denominator like that.
    ...
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  5. #5
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    Quote Originally Posted by skeeter View Post
    ...
    Okay, so how would I 'carry' on from there?

    Thanks
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  6. #6
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    Quote Originally Posted by Tweety View Post
    Okay, so how would I 'carry' on from there?

    Thanks
    $\displaystyle \sin^2 x + \cos^2 x =1,$ so $\displaystyle 1-\sin^2 x =$...
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  7. #7
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    Quote Originally Posted by Tweety View Post
    I dont really understand what you did, aren't you meant to prove that $\displaystyle c^{2} = \frac{16}{a^{2}-16} $? Why are you doing $\displaystyle c^{2} = cot^2{x} $?
    I started with $\displaystyle \frac{16}{a^2-16}$ and showed it was equal to $\displaystyle c^2$.

    note that one may start on either side of an equation and work to get the other side.
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