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C3 - Can somebody please explain this identity? Watch

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

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

    \equiv \dfrac{2\tan\theta}{1-\tan^2\theta}

    Now the bit I'm confused with is how to get from step 2 to 3. I know you divide by \cos^2\theta on both denominator and numerator.

    My question is how does this work? I'm on about in the case of \cos^2\theta = 0 In this sense you are in effect dividing by zero??

    I know this isn't the case and I'm being stupid, but I just want to know why :P Thanks
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    You wouldn't need to worry about that because whenever cos x = 0 then tan x is undefined: Wolfram
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    (Original post by Goldfishy)
    You wouldn't need to worry about that because whenever cos x = 0 then tan x is undefined: Wolfram
    Ahhh that makes sense

    How about in this identity then:

    \cos^2\theta + \sin^2\theta = 1 (divide by \sin^2\theta)

    \cot^2\theta + 1 = \mathrm{cosec}^2 \theta

    How does that work when  \sin\theta = 0 ??
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    (Original post by Quadratic)
    Ahhh that makes sense

    How about in this identity then:

    \cos^2\theta + \sin^2\theta = 1 (divide by \sin^2\theta)

    \cot^2\theta + 1 = \mathrm{cosec}^2 \theta

    How does that work when  \sin\theta = 0 ??
    Observe that  \displaystyle \frac{1}{sin\theta} = cosec\theta which you have used in your calculations. So when  sin\theta = 0, cosec\theta is undefined

    EDIT: Cheers for the rep, returned.
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    (Original post by Goldfishy)
    Observe that  \displaystyle \frac{1}{sin\theta} = cosec\theta which you have used in your calculations. So when  sin\theta = 0, cosec\theta is undefined

    EDIT: Cheers for the rep, returned.

    Oh aha Thank you so much for clearing that up for me

    And thank you for the rep!
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    (Original post by Quadratic)
    Oh aha Thank you so much for clearing that up for me

    And thank you for the rep!
    No problem
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    (Original post by Quadratic)
    Ahhh that makes sense

    How about in this identity then:

    \cos^2\theta + \sin^2\theta = 1 (divide by \sin^2\theta)

    \cot^2\theta + 1 = \mathrm{cosec}^2 \theta

    How does that work when  \sin\theta = 0 ??
    You're not actually carrying out the division, you're literally just rewriting it or changing the form.
 
 
 
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