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Math Help - proving expressions.

  1. #1
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    proving expressions.

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

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

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

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

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

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

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

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

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

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

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

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

     c = cotx , c^{2} = cot^{2}x

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

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

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

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

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

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

     a^{2} = 16 - 16cosec^2x

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

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

     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  c^{2} = \frac{16}{a^{2}-16} ? Why are you doing  c^{2} = cot^2{x} ?

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

     c = cotx , c^{2} = cot^{2}x

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

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

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

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

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

     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
    \sin^2 x + \cos^2 x =1, so 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  c^{2} = \frac{16}{a^{2}-16} ? Why are you doing  c^{2} = cot^2{x} ?
    I started with \frac{16}{a^2-16} and showed it was equal to c^2.

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