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Thread: Trig equation

  1. #1
    Senior Member Twig's Avatar
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    Trig equation

    Hi

    $\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta) $

    How do I solve these now again?

    Thx!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Twig View Post
    Hi

    $\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta) $

    How do I solve these now again?

    Thx!
    use this

    $\displaystyle cos(A+B)=cosAcosB-sinAsinB $
    and

    $\displaystyle cos(A-B)=cosAcosB+sinAsinB$
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  3. #3
    Senior Member I-Think's Avatar
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    $\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta)$

    Compound Angle Formula

    $\displaystyle cos\frac{\pi}{4}cos\theta-sin\frac{\pi}{4}sin\theta=\frac{1}{2}(cos\frac{\pi }{4}cos\theta+sin\frac{\pi}{4}sin\theta$

    $\displaystyle \frac{cos\theta}{\sqrt{2}}-\frac{sin\theta}{\sqrt{2}}=\frac{1}{2}(
    \frac{cos\theta}{\sqrt{2}}+\frac{sin\theta}{\sqrt{ 2}})$

    $\displaystyle *\sqrt{2}: cos\theta-sin\theta=\frac{1}{2}(cos\theta+sin\theta)$
    $\displaystyle *2: 2cos\theta-2sin\theta=cos\theta+sin\theta$
    $\displaystyle cos\theta-3sin\theta=0$
    $\displaystyle cos\theta=3sin\theta$
    $\displaystyle cos^2\theta=9sin^2\theta$
    $\displaystyle 1=10sin^2\theta$
    $\displaystyle \sqrt{\frac{1}{10}}=sin\theta$

    Solve from there. I'm sleepy
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  4. #4
    Senior Member Twig's Avatar
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    thx

    oh of course!

    Thanks alot!
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  5. #5
    Senior Member Twig's Avatar
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    Hi I-Think

    I belive the easiest approach after writing out the compound angle formulas are: $\displaystyle tan(\theta)=\frac{1}{3} $
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by I-Think View Post
    $\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta)$

    Compound Angle Formula

    $\displaystyle cos\frac{\pi}{4}cos\theta-sin\frac{\pi}{4}sin\theta=\frac{1}{2}(cos\frac{\pi }{4}cos\theta+sin\frac{\pi}{4}sin\theta$


    $\displaystyle \frac{cos\theta}{\sqrt{2}}-\frac{sin\theta}{\sqrt{2}}=\frac{1}{2}(
    \frac{cos\theta}{\sqrt{2}}+\frac{sin\theta}{\sqrt{ 2}})$

    $\displaystyle *\sqrt{2}: cos\theta-sin\theta=\frac{1}{2}(cos\theta+sin\theta)$

    $\displaystyle *2: 2cos\theta-2sin\theta=cos\theta+sin\theta$


    $\displaystyle cos\theta-3sin\theta=0$


    $\displaystyle cos\theta=3sin\theta$


    $\displaystyle cos^2\theta=9sin^2\theta$


    $\displaystyle 1=10sin^2\theta$


    $\displaystyle \sqrt{\frac{1}{10}}=sin\theta$

    Solve from there. I'm sleepy
    make spaces between [tex][/ math] and the [tex][/ math] that makes it easy to read
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  7. #7
    Super Member flyingsquirrel's Avatar
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    A slightly different approach :
    Quote Originally Posted by Twig View Post
    $\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta) $

    How do I solve these now again?
    As $\displaystyle \cos\left(\frac{\pi}{4}-\theta\right)=\sin\left(\frac{\pi}{4}+\theta\right )$, the above equation can be written $\displaystyle \tan\left(\frac{\pi}{4}+\theta\right)=2$. Expanding the left hand side gives us
    $\displaystyle \frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\cdot \tan\theta}=2\Longleftrightarrow \frac{1+\tan\theta}{1-1\cdot \tan\theta}=2\Longleftrightarrow 1+\tan \theta=2-2\tan \theta. $
    Thus $\displaystyle \tan \theta=1/3$...
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