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Math Help - Proving Questions leading to 'Hence' solve for x

  1. #1
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    Smile Proving Questions leading to 'Hence' solve for x

    I have a question to ask, would anyone of you explain to me how to solve this?

    1. Prove that (1+cosθ+sin2θ)/(1-cos2θ+sin2θ)=cotθ

    I'm able to prove it but,,



    Then the following part asks me solve the equation for 0≤θ≤2π

    1+sin2θ=3cos2θ

    How would you go about doing this? I appreciate your help!
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  2. #2
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    Re: Proving Questions leading to 'Hence' solve for x

    Is the first part supposed to be (1 + cos(2\theta) +sin(2\theta)) for the numerator?

    If not, can you show me your proof

    For the second part, wait till someone more experienced comes along; however, while you're waiting, here is my attempt at it...

    Let t = tan(\theta)

    1 + sin(2\theta) = 3cos(2\theta)

    \equiv~ ~\frac{1+t^2}{1+t^2} + \frac{2t}{1+t^2} - \frac{3(1-t^2)}{1+t^2} = 0

    Solve for t...

    tan(\theta) = \frac{1}{2}

    or

    tan(\theta) = -1

    For tan(\theta) = -1:

    \theta = arctan(-1) = -\frac{\pi}{4} therefore V=\{\frac{3\pi}{4}, \frac{7\pi}{4}\} are some solutions. Not 100% on how to solve for 1/2.
    Last edited by terrorsquid; August 20th 2011 at 09:18 AM.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Proving Questions leading to 'Hence' solve for x

    Probably you made a mistake in the numerator because I think the identity has to be \frac{1+\cos(2x)+\sin(2x)}{1-\cos(2x)+\sin(2x)}=\cot(x) (like terrorsquid already noticed).

    You can prove it this way (I use x in stead of \theta):
    \frac{1+\cos(2x)+\sin(2x)}{1-\cos(2x)+\sin(2x)}=\frac{1+(2\cos^2(x)-1)+2\sin(x)\cos(x)}{1-(1-2\sin^2(x))+2\sin(x)\cos(x)}
    =\frac{2\cos^2(x)+2\sin(x)\cos(x)}{2\sin^2(x)+2\si  n(x) \cos(x)}
    =\frac{2\cos(x)[\cos(x)+\sin(x)]}{2\sin(x)[\sin(x)+\cos(x)]}=\frac{\cos(x)}{\sin(x)}=\cot(x)

    The method Terrorsquid used is very useful.
    For the other solution:
    \tan(\theta)=\frac{1}{2} that means:
    \theta=\arctan\left(\frac{1}{2}\right)+k\pi
    Last edited by Siron; August 20th 2011 at 10:05 AM.
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