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Math Help - Trig equation problem

  1. #1
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    Trig equation problem

    Hi, I'm strugglin gto find a suitable identity or collection of identies to get this into a suitable form. Any pointers would be greatly received.

     \2cos2\theta + 4\sin\theta\cos\theta = \sqrt{2}

    I dont seem to beable to get rid of the constant (so can't factorize). Or get it to be in just one variable...

    Thanks, F
    Last edited by FelixHelix; October 17th 2011 at 04:49 AM. Reason: 2 x Typos
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  2. #2
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    Re: Trig equation problem

    Quote Originally Posted by FelixHelix View Post
    Hi, I'm strugglin gto find a suitable identity or collection of identies to get this into a suitable form. Any pointers would be greatly received.

     \cos2\theta = 4\sin\theta\cos\theta = \sqrt{2}

    I dont seem to beable to get rid of the constant (so can't factorize). Or get it to be in just one variable...

    Thanks, F
    Are you sure one of those equals signs isn't meant to be a + or -?
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  3. #3
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    Re: Trig equation problem

    Thanks for pointing this out, actually I have 2 typos!!

    So the equation should read:
    2\cos2\theta + 4\sin\theta\cos\theta = \sqrt{2}
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  4. #4
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    Re: Trig equation problem

    Quote Originally Posted by FelixHelix View Post
    Thanks for pointing this out, actually I have 2 typos!!

    So the equation should read:
    2\cos2\theta + 4\sin\theta\cos\theta = \sqrt{2}

    Divide both sides by 2.

    2 sin(θ) cos(θ) = sin(2θ)
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  5. #5
    Member sbhatnagar's Avatar
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    Thumbs up Re: Trig equation problem

    your equation is 2\cos{2\theta}+4\sin{\theta}\cos{\theta}=\sqrt{2}

    Dividing both sides by 2



    Now \sin{2\theta}=2\sin{\theta}\cos{\theta} (this is a formula) Applying this...

    \cos{2\theta}+\sin{2\theta}=\frac{1}{\sqrt{2}}

    \Rightarrow (\cos{2\theta}+\sin{2\theta})^2=\frac{1}{2}

    \Rightarrow \cos^2{2\theta}+\sin^2{2\theta}+2\sin{2\theta}\cos  {2\theta}=\frac{1}{2}

    \Rightarrow 1+\sin{4\theta}=\frac{1}{2}

    \Rightarrow \sin{4\theta}=-\frac{1}{2}

    \Rightarrow \4\theta=-\frac{\pi}{6}
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  6. #6
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    Re: Trig equation problem

    Thanks for the posts. I'd tried another way but was unsure if this method was valid: My workings are:
     2\cos2\theta + 4\sin\theta\cos\theta = \sqrt{2}

    Get it to the form using identities...

     \cos2\theta + \sin2\theta = \frac1{\sqrt{2}}

    But thought I could use the form  R\cos\left(2\theta - \alpha\right) \equiv \2\cos2\theta + 4\sin\theta\cos\theta

    Working this through I get:  \sqrt{2}\cos \left(2\theta - \frac\pi{4}\right) = \frac1{\sqrt{2}}

    And then the general solutions on the range  0 <\theta<2\pi will be:
    \theta= \frac{7\pi}{24}, \frac{23\pi}{24},\frac{31\pi}{24},\frac{47\pi}{24}

    Is my manipulation not allowed ??

    Thanks in advance for your comments.

    F
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  7. #7
    Super Member Quacky's Avatar
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    Re: Trig equation problem

    That's a perfectly valid (and correct) approach, as far as I can see.
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  8. #8
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    Re: Trig equation problem

    Thanks for taking the time to look. I've just used sbhatnagar method and used the principle value and I get more solutions including the 4 I got with my method.

    Which way is correct? Have I mis-manipulated mine and hence lost other answers?

    Thanks
    F
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  9. #9
    Super Member Quacky's Avatar
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    Re: Trig equation problem

    Quote Originally Posted by FelixHelix View Post
    Thanks for taking the time to look. I've just used sbhatnagar method and used the principle value and I get more solutions including the 4 I got with my method.

    Which way is correct? Have I mis-manipulated mine and hence lost other answers?

    Thanks
    F
    Sbhatnagar's method involves squaring both sides. This can introduce extraneous solutions. You'd have to check all solutions in the original equation if using his method.
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