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Math Help - Trigonometry Proof

  1. #1
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    Trigonometry Proof

    Can someone please prove this equation:

    Cos x + Cos x = 2Sec x
    1 - Sin x 1 + Sin x
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  2. #2
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    Quote Originally Posted by GAVREED2 View Post
    Can someone please prove this equation:

    Cos x + Cos x = 2Sec x
    1 - Sin x 1 + Sin x
    to start, get a common denominator and combine the two fractions on the left side ...
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  3. #3
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    I'm not sure if i completed the question correctly

    Is the answer 2/cos x = 2/cos x
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  4. #4
    MHF Contributor Unknown008's Avatar
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    You should consider only one hand side and should end up with 2sec(x).

    \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)} = \frac{cos(x)(1+sin(x) + cos(x)(1-sin(x))}{(1-sin(x))(1+sin(x))}

    Expand, simplify.

    Ultimately, you should get;

    \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)} = 2sec(x)
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  5. #5
    Senior Member Educated's Avatar
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    Quote Originally Posted by GAVREED2 View Post
    I'm not sure if i completed the question correctly

    Is the answer 2/cos x = 2/cos x
    Yes, the answer is \frac{2}{cos(x)} but that is not how you solve a prove question.

    There is no point in saying 2/cos x = 2/cos x, because of course that equation is true, and what you have basicly done is rewritten 2sec(x) in the form 2/cos(x).

    You should start off on the left, ignoring the right hand side, having only the equation \frac{cos(x)}{1-sin(x)} + \frac{cos(x)}{1+sin(x)} and simplifying/rearranging it to match what is stated.

    You must show the steps of how you got that, without the answer of 2sec(x) in the equation:

    \dfrac{cos(x)}{1-sin(x)} + \dfrac{cos(x)}{1+sin(x)}


     = \dfrac{cos(x)(1+sin(x) + cos(x)(1-sin(x))}{(1-sin(x))(1+sin(x))}


     = \dfrac{cos(x) + cos(x)sin(x) + cos(x) - cos(x)sin(x)}{1-sin^2(x)}

    Continue simplifying/rearranging it to get 2sec(x).

    Hint: ( cos^2(x) + sin^2(x) = 1)
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