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Math Help - please help sin cos

  1. #1
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    please help sin cos

    sqrt 2sinx + sin2x=0 for all values of x.

    Thank you.
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  2. #2
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    I'll assume that was meant to say 'solve' for all values of x.

    Quote Originally Posted by Trigon View Post
    sqrt 2sinx + sin2x=0 for all values of x.
    We have \sqrt{2}\sin{x}+\sin{2x} = 0 \Rightarrow \sqrt{2}\sin{x}+2\cos{x}\sin{x} = 0 \Rightarrow \sin{x}\left(\sqrt{2}+2\cos{x}\right) = 0.
    Thus either  \sin{x} = 0 or 2\cos{x} = -\sqrt{2} \Rightarrow \cos{x} = -\frac{1}{\sqrt{2}}. Recall that \sin{0} = 0 and \cos{\frac{\pi}{4} = \frac{1}{\sqrt{2}}.
    Then we have \sin{x} = 0 \Rightarrow \sin{x} = \sin{0} \Rightarrow x = 2n\pi or (2n+1)\pi. We can simply write this as k\pi ,
    where k\in\mathbb{Z} or zero. Also \cos{x} = -\frac{1}{\sqrt{2}} \Rightarrow \cos{x} = -\cos\frac{\pi}{4} = \cos\left(\pi+\frac{\pi}{4}\right) \Rightarrow x = 2k\pi \pm \frac{5\pi}{4}.
    Last edited by mr fantastic; March 5th 2011 at 02:41 AM.
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  3. #3
    Junior Member cupid's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    where k\in\mathbb{Z} or zero.
    \mathbb{Z} includes 0

    Quote Originally Posted by TheCoffeeMachine View Post
    Also \cos{x} = -\frac{1}{\sqrt{2}} \Rightarrow \cos{x} = -\cos\frac{\pi}{4} = \cos\left(\pi+\frac{\pi}{4}\right) \Rightarrow x = 2k\pi \pm \frac{5\pi}{4}.
    Cant we also write cosx = -cos\frac{\pi}{4} \implies cosx = cos(\pi - \frac{\pi}{4})
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  4. #4
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    Quote Originally Posted by cupid View Post
    \mathbb{Z} includes 0
    You're right, of course.
    Cant we also write cosx = -cos\frac{\pi}{4} \implies cosx = cos(\pi - \frac{\pi}{4})
    We can. But the aggregate values given by 2n\pi\pm\frac{5\pi}{4} and 2n\pi\pm\frac{3\pi}{4} are the same, so it doesn't matter which one we write.
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