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Math Help - Trig Identites

  1. #1
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    Trig Identites

    Prove: secθ - cosθ = sinθcosθ


    My Work:

    I know that secθ = 1/cosθ.

    So, I replace sine theta with 1/cos theta.

    (1/cosθ) - cosθ = sinθcosθ

    On the left side, I simply apply the rules for subtraction of fractions.

    The left side becomes (1 - cos^2θ)/cosθ.

    The equation now looks like this:

    (1 - cos^2θ)/cosθ = sinθcosθ

    Where do I go from there?
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  2. #2
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    1-\cos^2\theta = \sin^2\theta
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  3. #3
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    It seems to me that the trigonometric identity isn't true.

    \sec \theta - \cos \theta \ne \sin \theta \cos \theta


    Is the trigonometric identity supposed to be : \sec \theta - \cos \theta = \sin \theta \tan \theta = \frac{\sin^2 x}{\cos x} ??


    And also if you are proving trigonometric identities, you shouldn't have the answer at the end.
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  4. #4
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    Here is some formulae:

    \tan A = \frac{\sin A}{\cos A}

    \sec A = \frac{1}{\cos A}

    cosec A = \frac{1}{\sin A}

    \cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A}

    \sin^2 \theta = 1 - \cos^a \theta

    \tan^2 \theta = \sec^2 \theta - 1

    \cot^2 \theta = cosec^2 \theta -1

    Hope this helped....
    Last edited by Yudhishm; October 20th 2010 at 07:05 AM.
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  5. #5
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    Quote Originally Posted by RTC1996 View Post
    Prove: secθ - cosθ = sinθcosθ


    My Work:

    I know that secθ = 1/cosθ.

    So, I replace sine theta with 1/cos theta.

    (1/cosθ) - cosθ = sinθcosθ

    On the left side, I simply apply the rules for subtraction of fractions.

    The left side becomes (1 - cos^2θ)/cosθ.

    The equation now looks like this:

    (1 - cos^2θ)/cosθ = sinθcosθ

    Where do I go from there?
    hello,
    it seems that u cant solve the following identities as for proving, u have to make LHS=RHS
    and u can test out by substituting a number using the calculator and if u try,the LHS =/= RHS.
    To prove the equation,the LHS must be = to RHS.
    correct me if im wrong.
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  6. #6
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    Quote Originally Posted by Educated View Post
    It seems to me that the trigonometric identity isn't true.

    \sec \theta - \cos \theta \ne \sin \theta \cos \theta


    Is the trigonometric identity supposed to be : \sec \theta - \cos \theta = \sin \theta \tan \theta = \frac{\sin^2 x}{\cos x} ??


    And also if you are proving trigonometric identities, you shouldn't have the answer at the end.

    I agree that this trig problem has no solution.
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  7. #7
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    I don't think there is a solution in this case.
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  8. #8
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    Quote Originally Posted by pickslides View Post
    1-\cos^2\theta = \sin^2\theta
    I don't think there is a solution in this case.
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