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Thread: Explain working

  1. #1
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    Smile Explain working

    Seek explanation of getting the second line, $\displaystyle = 8(\pi+ \frac{\pi}{2}) - (0)$.

    $\displaystyle A=8 \left [\theta + 2sin\theta + \frac{\theta}{2} + \frac{sin2\theta}{4} \right ]^\pi_0$

    $\displaystyle = 8(\pi+ \frac{\pi}{2}) - (0)$

    $\displaystyle =8 \pi + 4 \pi = 12\pi $ Ans


    I know that $\displaystyle sin2 \theta = 2sin \theta cos \theta$

    so that,

    $\displaystyle 2sin \theta + \frac{sin2 \theta}{4} = 2sin\theta + \frac{2sin \theta cos \theta}{4}$

    $\displaystyle =2sin \theta \left ( 1 + \frac {cos \theta}{4} \right )$

    $\displaystyle ............................$

    $\displaystyle ............................$
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  2. #2
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    Quote Originally Posted by ashura View Post
    Seek explanation of getting the second line, $\displaystyle = 8(\pi+ \frac{\pi}{2}) - (0)$.

    $\displaystyle A=8 \left [\theta + 2sin\theta + \frac{\theta}{2} + \frac{sin2\theta}{4} \right ]^\pi_0$

    $\displaystyle = 8(\pi+ \frac{\pi}{2}) - (0)$

    $\displaystyle =8 \pi + 4 \pi = 12\pi $ Ans


    I know that $\displaystyle sin2 \theta = 2sin \theta cos \theta$

    so that,

    $\displaystyle 2sin \theta + \frac{sin2 \theta}{4} = 2sin\theta + \frac{2sin \theta cos \theta}{4}$

    $\displaystyle =2sin \theta \left ( 1 + \frac {cos \theta}{4} \right )$
    Hello, Ashura,

    you only have to remember that $\displaystyle \sin(0)=\sin(\pi)=\sin(2\cdot\pi)=...=0$

    thus when you plug in pi all summands containing the sine function will become zero and when you plug in zero even the theta-summands will become zero too.

    EB
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  3. #3
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    Wink

    Thanks, I understand the 0 result when 0 is pluged in for theta. I'm not seeing how the $\displaystyle \left (\pi + \frac {\pi}{2} \right )$ result is reached when pi is plugged in for theta. That's what I was trying to work out in the last 2 lines of my post.
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  4. #4
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    Quote Originally Posted by ashura View Post
    $\displaystyle 8 \left [\theta + 2sin\theta + \frac{\theta}{2} + \frac{sin2\theta}{4} \right ]$ for $\displaystyle \theta = \pi$
    $\displaystyle 8 \left [\pi + 2sin\pi + \frac{\pi}{2} + \frac{sin2\pi}{4} \right ]$

    As Earboth mentioned:
    $\displaystyle sin\pi = 0$
    $\displaystyle sin2\pi = 0$

    So:
    $\displaystyle 8 \left [\pi + 0 + \frac{\pi}{2} + \frac{0}{4} \right ]$

    $\displaystyle 8 \left [\pi + \frac{\pi}{2} \right ]$

    -Dan
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  5. #5
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    Thanks, didn't realize that $\displaystyle sin \, \pi $ was really 0.
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