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Thread: Trigonometry

  1. #1
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    Trigonometry

    Given that $\displaystyle cos 3x = 4cos^3x -3cosx$,

    (a) Find the exact value of $\displaystyle 8cos^310 - 6cos10$
    (b) solve the equation $\displaystyle 8cos^3x - 6cosx = - 1$ for 0< x < 180

    Attempted solution

    (a) 8cos^310 - 6cos10 = 2(cos3(10))
    = 2cos30
    = $\displaystyle \sqrt{3}$

    (b) since 8cos^3x - 6cosx = 2cos 3x
    2cos3x = -1
    I cosine inversed $\displaystyle \frac{-1}{2}$ but am not clear on what to do after that.
    I know about the positive and negative quadrants.

    However, the 3x is giving me some problems on adding and minus...

    Note that 0< x <180
    Last edited by Punch; Jan 9th 2010 at 04:00 AM.
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  2. #2
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    Quote Originally Posted by Punch View Post
    Given that $\displaystyle cos 3x = 4cos^3x -3cosx$,

    (a) Find the exact value of $\displaystyle 8cos^310 - 6cos10$
    (b) solve the equation $\displaystyle 8cos^3x - 6cosx = - 1$ for 0< x < 180

    Attempted solution

    (a) 8cos^310 - 6cos10 = 2(cos3(10))
    = 2cos30
    = $\displaystyle \sqrt{3}$

    (b) since 8cos^3x - 6cosx = 2cos 3x
    2cos3x = -1
    I cosine inversed $\displaystyle \frac{-1}{2}$ but am not clear on what to do after that.
    I know about the positive and negative quadrants.

    However, the 3x is giving me some problems on adding and minus...

    Note that 0< x <180
    Do not double post.

    Question has been answered in the other thread.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Do not double post.

    Question has been answered in the other thread.
    I am sorry this always happens because of some bugs i have in my comp..
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  4. #4
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    Hello Punch
    Quote Originally Posted by Punch View Post
    Given that $\displaystyle cos 3x = 4cos^3x -3cosx$,

    (a) Find the exact value of $\displaystyle 8cos^310 - 6cos10$
    (b) solve the equation $\displaystyle 8cos^3x - 6cosx = - 1$ for 0< x < 180

    Attempted solution

    (a) 8cos^310 - 6cos10 = 2(cos3(10))
    = 2cos30
    = $\displaystyle \sqrt{3}$

    (b) since 8cos^3x - 6cosx = 2cos 3x
    2cos3x = -1
    I cosine inversed $\displaystyle \frac{-1}{2}$ but am not clear on what to do after that.
    I know about the positive and negative quadrants.

    However, the 3x is giving me some problems on adding and minus...

    Note that 0< x <180
    Thanks for showing us your working so far. Your answer to (a) is perfect.

    For (b), you're right so far. Note that if $\displaystyle 0^o<x<180^o$, then $\displaystyle 0^o<3x<540^o$. Then you can say, since $\displaystyle \cos3x = -\tfrac12$, we want to find all angles between $\displaystyle 0^o$ and $\displaystyle 540^o$ whose cosine is $\displaystyle -\tfrac12$.

    The first of these is $\displaystyle (180-60)=120^o$, the next is $\displaystyle (180+60)=240^o$, and so on. And these are the values of $\displaystyle 3x$. So we get:
    $\displaystyle 3x = 120^o, 240^o, 480^o, 600^o, ...$ (the last one being out of range.)

    $\displaystyle \Rightarrow x = 40^o, 80^o, 160^o$
    Grandad
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