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Math Help - Trigonometric equations

  1. #1
    Member roshanhero's Avatar
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    Trigonometric equations

    In my book following result is shown but don't know where it come from.
    1.2cos^2(2kpie)=2 (where k is an integer)
    2.cos(kpie)=(-1)^k
    3.[sin(kpie)cos(theta)-cos(kpie)sin(theta)]=[(-1)^k-1 sin(theta)]
    In all these cases I just don't know how the results came.Please explain it to me.
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  2. #2
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    Hi roshanhero,

    1) The cosine function is periodic with period 2\pi and thus \cos (2k\pi)=\cos (2\pi)~\forall k \in \mathbb{Z} , hence 2\cos^2 (2k\pi)=2\cos^2 (2\pi)~\forall k \in \mathbb{Z} The conclusion follows.

    2) If k is even then k=2n for some n and hence \cos (k\pi)=\cos (2n\pi)=\cos (2\pi)~\forall n \in \mathbb{Z} (which follows from above). Notice that (-1)^k=[(-1)^2]^n=1^n=1~\forall n \in \mathbb{Z} Thus if k is even then the statement is true since \cos (2\pi)=1 Note that due to the cosine function's periodic attributes \cos (\psi+\pi)\equiv-\cos (\psi) . If k is odd then k=2r+1 for some r and hence \cos (k\pi)=\cos [(2r+1)\pi]=\cos (2r\pi+\pi)=-\cos (2r\pi) = -\cos (2\pi)~~\forall r \in \mathbb{Z} Notice that (-1)^k=(-1)^{2r+1}=(-1)\cdot(-1)^{2r}=-1 Thus if k is odd then the statement is true since -\cos (2\pi)=-1 hence the statement is true \forall k \in \mathbb{Z}

    3) I have to go so I will leave this one until later unless of course someone else would like to fill in the blanks.
    Last edited by Sean12345; August 23rd 2008 at 10:47 AM.
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  3. #3
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    I'm back.

    3) Note that \sin (\theta \pm \phi) \equiv \sin (\theta) \cdot \cos (\phi) \pm \cos (\theta) \cdot \sin (\phi) Following from this we can see that \sin (k\pi) \cdot \cos (\theta) - \cos (k\pi) \cdot sin (\theta) \equiv \sin (k\pi - \theta) Like previously we will split this into two seperate cases. If k is even then k=2n for some n and hence \sin (k\pi - \theta) = \sin (-\theta + 2n\pi) The sine function is also periodic with period 2\pi and thus \sin(\phi + 2x\pi) \equiv \sin (\phi)~ \forall x \in \mathbb{Z} Also note that \sin(-\psi) \equiv -\sin (\psi) Knowing these two bits of information we can then see that \sin (-\theta + 2n\pi) \equiv \sin (-\theta) \equiv -\sin (\theta) Notice that (-1)^{k-1} \cdot \sin (\theta) = \frac{(-1)^{2n}}{-1} \cdot \sin (\theta) = -\sin (\theta)~\forall n \in \mathbb{Z} Hence the statement is true if k is even since rather trivially -\sin (\theta) = -\sin (\theta) .

    Now if k is odd then k=2r+1 for some r and thus \sin (k\pi - \theta) = \sin [-\theta + (2n+1) \pi] = \sin (-\theta + 2n \pi + \pi) = \sin (-\theta + \pi) Note that the sine function has some of the same attributes as the cosine function in that \sin (\psi +\pi) = -\sin (\psi) Hence using this results implies that \sin (-\theta + \pi) = -\sin (-\theta) = \sin (\theta) Notice that (-1)^{k-1} \cdot \sin (\theta) = (-1)^{2r} \cdot \sin (\theta) = \sin (\theta)~\forall r \in \mathbb {Z} Hence the statement is true if k is odd since again rather trivially \sin (\theta) \equiv \sin (\theta) and thus the statement is true ~\forall k \in \mathbb {Z}

     Q.E.D.

    Thats how the results came.
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