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Math Help - Trig eqn cos(x+TT/6) + cos(x-TT/4) = 0

  1. #1
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    Trig eqn cos(x+TT/6) + cos(x-TT/4) = 0

    Hi there, I was wondering if someone would be able to help me solve this equation :

    cos(x+TT/6) + cos(x-TT/4) = 0

    It's driving me insane. I use compound angle theorem to expand the components and end up with:
    \sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}+1/\sqrt{2}sin{x}=0

    I then tried squaring both sides and using trig identities to put everything in terms of cos^2{x}. And got something that I wont waste your time with. Please help!
    Last edited by flashylightsmeow; February 25th 2011 at 12:35 PM. Reason: I made an error with one of the sign of the last term it should have been positive.
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  2. #2
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    Quote Originally Posted by flashylightsmeow View Post
    \sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}-1/\sqrt{2}sin{x}=0

    now group some like terms, what do you get?
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  3. #3
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    Quote Originally Posted by flashylightsmeow View Post
    Hi there, I was wondering if someone would be able to help me solve this equation :

    cos(x+TT/6) + cos(x-TT/4) = 0

    It's driving me insane. I use compound angle theorem to expand the components and end up with:
    \sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}-1/\sqrt{2}sin{x}=0

    I then tried squaring both sides and using trig identities to put everything in terms of cos^2{x}. And got something that I wont waste your time with. Please help!
    Since the right hand side is zero why not use the sum to product identities.

    \displaystyle \cos(a)+\cos(b)=2\cos\left( \frac{a+b}{2}\right)\cos\left( \frac{a-b}{2}\right)

    This gives

    2\displaystyle \cos\left( \frac{2x-\frac{\pi}{12}}{2}\right)\cos\left( \frac{\frac{5\pi}{12}}{2}\right)=0
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  4. #4
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    Quote Originally Posted by pickslides View Post
    now group some like terms, what do you get?
    So i squared everything and multiplied through by 4 to get rid of the denominators and got

    6cos^2{x}-2sin^2{x}+4cos^2{x}+4sin^2{x}=0

    then grouped together like terms:


    10cos^2{x}+2sin^2{x}=0

    but from here
    10cos^2{x}+2sin^2{x}=0


    then: using the pythagorean trig identity

    10cos^2{x}+2(1-cos^2{x})=0

    8cos^2{x}+2=0

    but from here:
    cos^2{x}=-2/8 which doesn't give a real number AGH!!!!
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  5. #5
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    ah i didn't even look at it like that. I'll give it a shot, and if it doesnt work, I'll be crawling back!
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  6. #6
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    Quote Originally Posted by pickslides View Post
    now group some like terms, what do you get?
    Hi Pickslides, I made an error with the sign on the last term, it should be a + instead of -.
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    Since the right hand side is zero why not use the sum to product identities.

    \displaystyle \cos(a)+\cos(b)=2\cos\left( \frac{a+b}{2}\right)\cos\left( \frac{a-b}{2}\right)

    This gives

    2\displaystyle \cos\left( \frac{2x-\frac{\pi}{12}}{2}\right)\cos\left( \frac{\frac{5\pi}{12}}{2}\right)=0
    so from this i get:

    2cos(x-{\pi/24})cos({5\pi/24})=0

    Can i deduce from this expression that cos(x-{\pi/24}) must equal 0?

    if so where do I go from there?
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    Quote Originally Posted by flashylightsmeow View Post
    so from this i get:

    2cos(x-{\pi/24})cos({5\pi/24})=0

    Can i deduce from this expression that cos(x-{\frac{\pi}{24}}) must equal 0?

    if so where do I go from there?
    Yes then

    \cos(x-{\frac{\pi}{24}})=0 \iff x-{\frac{\pi}{24}}=\cos^{-1}(0)=\frac{\pi}{2}+n\pi for n \in \mathbb{Z}
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  9. #9
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    OH yea! Man I am bad at maths. Just a quick question, is it possible to use the method i initially used using cos(x+y) - cos(x-y)? I don't get where i went wrong :S
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  10. #10
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    Solution by exploiting symmetry

    Quote Originally Posted by flashylightsmeow View Post
    \cos\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0
    By \cos\left(\pi+\theta \right) = -\cos{\theta}, we have \cos\left(x-\frac{\pi}{4}\right) = \cos\left(\pi+x-\frac{\pi}{4}-\pi\right) = -\cos\left(x-\frac{5\pi}{4}\right).

    Thus \cos\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0 \Rightarrow \cos\left(x+\frac{\pi}{6}\right)  = -\cos\left(x-\frac{\pi}{4}\right) = \cos\left(x-\frac{5\pi}{4}\right).

    So \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right). If \cos{\theta} = \cos{\alpha}, then \theta = 2n\pi + \alpha or \theta = 2n\pi -\alpha, thus:

    \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi + x - \frac{5\pi}{4} \Rightarrow 0 = 2n\pi-\frac{17\pi}{12}{\pi}~ (dead-end).

    Or we have \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi -x + \frac{5\pi}{4} \Rightarrow x = n\pi+\frac{13\pi}{24} ~  \checkmark.

    Most problems of this kind can be solved by some consideration of symmetry; a twin of your problem is:

    \sin\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0
    \cos{\theta} = \sin\left(\frac{\pi}{2}-\theta\right), thus \cos\left(x-\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}+\frac{\pi}{4}-x\right) =  \sin\left(\frac{3\pi}{4}-x\right), hence

     \sin\left(x+\frac{\pi}{6}\right)  = -\cos\left(x-\frac{\pi}{4}\right) \Rightarrow \sin\left(x+\frac{\pi}{6}\right)  = -\sin\left(\frac{3\pi}{4}-x\right) = \sin\left(x-\frac{3\pi}{4}\right).

    So \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right). If \sin{\theta} = \sin{\alpha}, then \theta = 2n\pi+\alpha or (2n+1)\pi-\alpha, thus

    \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi+x-\frac{3\pi}{4} \Rightarrow 0 = 2n\pi-\frac{3\pi}{4}-\frac{\pi}{6} (dead-end).

    Or \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = (2n+1)\pi-x+\frac{3\pi}{4} \Rightarrow x = (2n+1)\frac{\pi}{2}+\frac{7\pi}{24} ~ \checkmark.
    Last edited by TheCoffeeMachine; February 25th 2011 at 04:42 PM.
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