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Math Help - Please check proof of identity

  1. #1
    Member Furyan's Avatar
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    Please check proof of identity

    Hello all

    The question is show that:

    \dfrac{\sin 2\theta}{1+\cos 2\theta} \equiv \tan\theta

    I have

    LHS \equiv\dfrac{\sin\theta\cos\theta + \cos\theta\sin\theta}{1 + \cos^2\theta - \sin^2\theta}

    \equiv\dfrac{2\sin\theta\cos\theta}{1 + \cos^2\theta - (1 - cos^2\theta)}

    \equiv\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}

    \equiv\dfrac{\sin\theta}{\cos\theta} \equiv \tan\theta

    Is that alright and is the a better way?

    Thank you.
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  2. #2
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    e^(i*pi)'s Avatar
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    Re: Please check proof of identity

    Quote Originally Posted by Furyan View Post
    Hello all

    The question is show that:

    \dfrac{\sin 2\theta}{1+\cos 2\theta} \equiv \tan\theta

    I have

    LHS \equiv\dfrac{\sin\theta\cos\theta + \cos\theta\sin\theta}{1 + \cos^2\theta - \sin^2\theta}

    \equiv\dfrac{2\sin\theta\cos\theta}{1 + \cos^2\theta - (1 - cos^2\theta)}

    \equiv\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}

    \equiv\dfrac{\sin\theta}{\cos\theta} \equiv \tan\theta

    Is that alright and is the a better way?

    Thank you.
    That way is fine. I'd have omitted the third line of working since the reader should be able to establish that you converted the existing 1-\sin^2 \theta into \cos^2 \theta

    Spoiler:
    This one -- \equiv\dfrac{2\sin\theta\cos\theta}{1 + \cos^2\theta - (1 - cos^2\theta)}
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  3. #3
    Member Furyan's Avatar
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    Re: Please check proof of identity

    Quote Originally Posted by e^(i*pi) View Post
    That way is fine. I'd have omitted the third line of working since the reader should be able to establish that you converted the existing 1-\sin^2 \theta into \cos^2 \theta

    Spoiler:
    This one -- \equiv\dfrac{2\sin\theta\cos\theta}{1 + \cos^2\theta - (1 - cos^2\theta)}
    Thank you very much indeed, I didn't see that. I keep missing simple things like that. Thank you for checking my work and pointing out that step.
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