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Thread: General Solution to sin function

  1. #1
    Member alexgeek's Avatar
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    Question General Solution to sin function

    I have a question in my further maths book to find the general solution in radians of:
    $\displaystyle
    \sin(2\theta + \frac{\pi}{4} ) = 1
    $
    The textbook says the answer is:
    $\displaystyle n\pi + \frac{\pi}{8}$

    Though this is what I did and got a different answer:
    $\displaystyle \alpha = \arcsin(1) = \frac{\pi}{2} $
    Put it into the general solution formula:
    $\displaystyle 2\theta + \frac{\pi}{4} = n\pi +(-1)^n \frac{\pi}{2} $
    Subtracted:
    $\displaystyle 2\theta = n\pi +(-1)^n \frac{\pi}{2} - \frac{\pi}{4} $
    and then divided to get:
    $\displaystyle \theta = n \frac{\pi}{2} + (-1)^n \frac{\pi}{4} - \frac{\pi}{8} $

    Have I gone wrong or is there a misprint?
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  2. #2
    MHF Contributor red_dog's Avatar
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    No, you're right.

    But if $\displaystyle n=2p$, then $\displaystyle \theta=p\pi+\frac{\pi}{4}-\frac{\pi}{8}=p\pi+\frac{\pi}{8}$

    and if $\displaystyle n=2p+1$, then $\displaystyle \theta=p\pi+\frac{\pi}{2}-\frac{\pi}{4}-\frac{\pi}{8}=p\pi+\frac{\pi}{8}$

    So, $\displaystyle \theta=k\pi+\frac{\pi}{8}, \ k\in\mathbb{Z}$
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  3. #3
    Member alexgeek's Avatar
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    Sorry, what does p represent?
    And how does $\displaystyle (-1)^n$ disappear?
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by alexgeek View Post
    Sorry, what does p represent?
    And how does $\displaystyle (-1)^n$ disappear?
    p is an integer.

    n can be even or odd. If n is even, then n=2p and $\displaystyle (-1)^n=(-1)^{2p}=1$.

    If n is odd then n=2p+1 and $\displaystyle (-1)^n=(-1)^{2p+1}=-1$
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  5. #5
    Member alexgeek's Avatar
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    Oh I see you split it into two formulas to get rid of the plus/minus.
    Thankyou!
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