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Math Help - Proving and then solving eqns

  1. #1
    Member helloying's Avatar
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    Proving and then solving eqns

    my question asked me to prove that \cos 3\theta= 4\cos^3\theta - 3\cos\theta

    i proved already and then the next part asked me to substitute x=6\cos\theta to find the roots of the eqn x^3 - 27x + 8=0

    i dont know what is point of substituting it. i thought i can just find the roots by keying the eqn into my calculator. can people pls teach me how to find the roots of the eqn without using the calculator?
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  2. #2
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    Quote Originally Posted by helloying View Post
    my question asked me to prove that \cos 3\theta= 4\cos^3\theta - 3\cos\theta

    i proved already and then the next part asked me to substitute x=6\cos\theta to find the roots of the eqn x^3 - 27x + 8=0

    i dont know what is point of substituting it. i thought i can just find the roots by keying the eqn into my calculator. can people pls teach me how to find the roots of the eqn without using the calculator?
    The point is to make use of the identity above.

    (6\cos\theta)^3 - 27(6\cos\theta) + 8 = 0

    216{\cos}^3 \theta - 162\cos\theta + 8 = 0

    54(4{\cos}^3 \theta - 3\cos\theta) + 8 = 0

    54\cos 3\theta + 8 = 0

    54\cos 3\theta = -8

    \cos 3\theta = -\frac{8}{54}

    \cos 3\theta = -\frac{4}{27}

    At this point you can use the calculator to find \theta. Then plug each answer into x=6\cos\theta to find x.


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  3. #3
    Member great_math's Avatar
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    (6\cos\theta)^3 - 27(6\cos\theta) + 8 = 0

    \Rightarrow 216{\cos}^3 \theta - 162\cos\theta + 8 = 0

    \Rightarrow 54\cos 3\theta + 8 = 0

    \Rightarrow \cos 3\theta = -\frac{8}{54}

    \Rightarrow 3\theta=\cos^{-1}\left(\frac{-4}{27}\right)
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  4. #4
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    Forgot to mention, when you try to solve
    \cos 3\theta = -\frac{4}{27}
    You should get six different values for \theta in 0 \le \theta < 2\pi, but when you plug each value into x = 6\cos \theta you only get three different values, which makes sense, because the original equation is a cubic.


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  5. #5
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    Hi,
    I've also been given this question, but I'm struggling proving the identity.
    Any help would be gratefully received.
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  6. #6
    Member great_math's Avatar
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    Quote Originally Posted by ady72 View Post
    Hi,
    I've also been given this question, but I'm struggling proving the identity.
    Any help would be gratefully received.
    \cos 3\theta=\cos(2\theta+\theta)

    =\cos 2\theta\cdot\cos\theta-\sin2\theta\cdot\sin\theta

    =(2\cos^2\theta-1)\cos\theta-2\sin^2\theta\cos\theta

    =2\cos^3\theta-\cos\theta-2(1-\cos^2\theta)\cos\theta

    =2\cos^3\theta-\cos\theta-2\cos\theta+2\cos^3\theta

    =4\cos^3\theta-3\cos\theta
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  7. #7
    Member helloying's Avatar
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    Quote Originally Posted by ady72 View Post
    Hi,
    I've also been given this question, but I'm struggling proving the identity.
    Any help would be gratefully received.
    \cos3\theta=\cos(2\theta+\theta)=\underline{\cos2\  theta}\cos\theta-\sin2\theta\sin\theta
    \underline{\cos2\theta}=\cos(\theta+\theta)=\cos\t  heta\cos\theta-\sin\theta\sin\theta

    i given u the 1st step and important step. try to simplify it
    Last edited by helloying; June 6th 2009 at 06:27 AM.
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  8. #8
    Member helloying's Avatar
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    Quote Originally Posted by yeongil View Post
    Forgot to mention, when you try to solve
    \cos 3\theta = -\frac{4}{27}
    You should get six different values for \theta in 0 \le \theta < 2\pi, but when you plug each value into x = 6\cos \theta you only get three different values, which makes sense, because the original equation is a cubic.


    01
    thanks for ur reply. i understand now but i am not sure how to get 3 values.i did by finding cosine is negative so angle is in the 2nd and 3rd quardant. then i find the reference angle and i got the angles in the 2nd and 3rd quardants. but that is only two ans. how to find the 3rd one?
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  9. #9
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    Quote Originally Posted by helloying View Post
    thanks for ur reply. i understand now but i am not sure how to get 3 values.i did by finding cosine is negative so angle is in the 2nd and 3rd quardant. then i find the reference angle and i got the angles in the 2nd and 3rd quardants. but that is only two ans. how to find the 3rd one?
    You got the angles in Quadrants 2 and 3, but they were 3\theta, not \theta. Think of it this way. If we are limiting \theta to 0 \le \theta < 2\pi, then we need to limit 3\theta to 0 \le 3\theta < 6\pi.

    I'm assuming that these were two of the angles you found:
    3\theta \approx 98.5\ deg \rightarrow \theta \approx 32.8\ deg

    3\theta \approx 261.5\ deg \rightarrow \theta \approx 87.2\ deg

    (Sorry that I'm mixing degrees and radians here.) To find the remaining angles, you'll have to add 360 degrees, then 720 degrees to each of the angles above, before dividing by 3. That way, you'll find all values of theta between 0 and 360 degrees. So the third angle is:

    3\theta \approx 98.5 + 360 \approx 458.5\ deg \rightarrow \theta \approx 152.8\ deg

    The sixth and final angle should be:

    3\theta \approx 261.5 + 720 \approx 981.5\ deg \rightarrow \theta \approx 327.2\ deg

    Now plug in each angle into x = 6\cos \theta, and as I said before, you'll only get three different values.


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