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Math Help - Finding cos2*theta

  1. #1
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    Finding cos2*theta

    Hey,
    I have a problem that I couldn't solve, I have no idea of the method I shuold use to solve it, please anyone who does can you show me?

    The question is attached...

    Thanks
    Attached Thumbnails Attached Thumbnails Finding cos2*theta-math.jpg  
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  2. #2
    Senior Member BAdhi's Avatar
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    Re: Finding cos2*theta

    draw a rectangled triangle and mark an angle as 2\theta and mark appropriate lengths of the sides which will give \sin{2\theta}=\frac{4\sqrt{5}}{9} and find the unknown side so that you can find \cos{2\theta}. this is same as using the identity \cos^2{\theta}=1-\sin^2{\theta}
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  3. #3
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    Re: Finding cos2*theta

    Quote Originally Posted by BAdhi View Post
    draw a rectangled triangle and mark an angle as 2\theta and mark appropriate lengths of the sides which will give \sin{2\theta}=\frac{4\sqrt{5}}{9} and find the unknown side so that you can find \cos{2\theta}. this is same as using the identity \cos^2{\theta}=1-\sin^2{\theta}
    I'm sorry, but I don't understand you wrote....
    Thanks for the attempt though :-)
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  4. #4
    Senior Member BAdhi's Avatar
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    Re: Finding cos2*theta

    either use the identity \cos^2{2\theta}=1-\sin^2{2\theta} or

    use the following triangle to find \cos{2\theta}
    Attached Thumbnails Attached Thumbnails Finding cos2*theta-untitled.jpg  
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  5. #5
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    Re: Finding cos2*theta

    I tried your triangle method and got 2/9, is this correct?
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  6. #6
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    Re: Finding cos2*theta

    Quote Originally Posted by IBstudent View Post
    I tried your triangle method and got 2/9, is this correct?
    No this isn't correct, but it's close. The identity that BAdhi posted is one of the most useful in Trigonometry, and really should be memorized. Knowing it will make much of the computation you end up doing so much easier.
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  7. #7
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    Re: Finding cos2*theta

    Hello, IBstudent!

    \sin2\theta \:=\:\frac{4\sqrt{5}}{9}

    \text{Find the exact value of }\cos2\theta.

    \text{Recall: }\:\sin^2\!x + \cos^2\!x \:=\:1


    So we have: . \sin^2\!2\theta + \cos^2\!2\theta \:=\:1

    . . . . . . . . \left(\frac{4\sqrt{5}}{9}\right)^2 + \cos^2\!2\theta \:=\:1

    . . . . . . . . . . . . \frac{80}{81} + \cos^2\!2\theta \:=\:1

    . . . . . . . . . . . . . . . . \cos^2\!2\theta \:=\:\frac{1}{81}

    . . . . . . . . . . . . . . . . \cos2\theta \:=\:\pm\frac{1}{9}

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  8. #8
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    Re: Finding cos2*theta

    Quote Originally Posted by IBstudent View Post
    Hey,
    I have a problem that I couldn't solve, I have no idea of the method I shuold use to solve it, please anyone who does can you show me?

    The question is attached...

    Thanks
    Are you told which quadrant of the unit circle that the angle \displaystyle 2\theta lies in?
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