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Math Help - How to solve a trigometric function...

  1. #1
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    How to solve a trigometric function...

    Hey everybody.
    I got this math assignment and I've been trying to work this problem but I have NO idea how.

    Can anyone help?:
    [tex]squareroot[1-cos2theta] = 2sinsqaredtheta/MATH]

    I apologize in advance if the math button didn't work...
    I'm still new to this forum!

    Any help is appreciated,
    Stealth
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  2. #2
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    Hello, Stealth!

    I'll make up my own instructions . . .


    Solve for 0 \leq x < 2\pi\!:\quad \sqrt{1-\cos2\theta} \:=\:2\sin^2\theta
    We're expected to know the identity: . \sin^2\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 1-\cos2\theta \:=\:2\sin^2\!\theta

    So the left side is: . \sqrt{1-\cos2\theta} \:=\:\sqrt{2\sin^2\!\theta} \:=\:\sqrt{2}\,\sin\theta


    The equation becomes: . \sqrt{2}\,\sin\theta \:=\:2\sin^2\!\theta \quad\Rightarrow\quad 2\sin^2\!\theta - \sqrt{2}\,\sin\theta \:=\:0

    Factor: . \sin\theta\,(2\sin\theta - \sqrt{2}) \:=\:0


    And we have: . \begin{array}{ccccccc}\sin\theta \:=\:0 & \Rightarrow & \boxed{\theta \:=\:0,\pi} \\ \\[-4mm]<br />
2\sin\theta - \sqrt{2}\:=\:0 & \Rightarrow & \sin\theta \:=\:\dfrac{\sqrt{2}}{2} & \Rightarrow &\boxed{ \theta \:=\:\dfrac{\pi}{4},\:\dfrac{3\pi}{4}} \end{array}

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  3. #3
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    Wow! Thanks! I was looking at the sinsquared formula but I didn't know that I could manipulate the 2 to make it 2sinsquaredtheta = 1-cos2theta!

    Thank you sooo much!

    Stealth
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  4. #4
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    Edit

    Hi! I noticed something when I worked it out another way!:


    We're expected to know the identity: . \sin^2\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 1-\cos2\theta \:=\:2\sin^2\!\theta

    So the left side is: . \sqrt{1-\cos2\theta} \:=\:\sqrt{2\sin^2\!\theta} \:=\:\sqrt{2}\,\sin\theta

    On the squareroot(2) part, would that be a plus/minus squareroot(2) because then wouldn't we also have the solutions of 5pi/4, and 7pi/4?

    The equation becomes: . \sqrt{2}\,\sin\theta \:=\:2\sin^2\!\theta \quad\Rightarrow\quad 2\sin^2\!\theta - \sqrt{2}\,\sin\theta \:=\:0
    [/quote]
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