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Thread: Trigonometric Functions

  1. #1
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    Trigonometric Functions

    The final line of my attempt contains the 2 answers listed in the book, but I'm not sure what the basis for excluding all the other values under $\displaystyle 180^{\circ}$ is? I have a feeling my approach is slightly incorrect here. Can anyone help me clear this up?

    Many thanks.

    Q.
    Solve the equation $\displaystyle \sin2x=\frac{1}{\sqrt{2}}$ for $\displaystyle 0\leq x\leq\pi$.

    Attempt: $\displaystyle \sin2x=\frac{1}{\sqrt{2}}\rightarrow2x=45^{\circ}$. Sin is positive in the 1st & 2nd quadrants.
    $\displaystyle \sin2x=\frac{1}{\sqrt{2}}\rightarrow2x=0^{\circ}+4 5^{\circ}=45^{\circ}$ or $\displaystyle 180^{\circ}-45^{\circ}=135^{\circ}$
    Since the angle is $\displaystyle 2x$ add $\displaystyle 90^{\circ}$ & $\displaystyle 270^{\circ}$ to each angle (i.e. $\displaystyle 2(45^{\circ})$ & $\displaystyle 2(135^{\circ}$):
    $\displaystyle 2x=45^{\circ},135^{\circ},315^{\circ}$ or $\displaystyle 2x=135^{\circ},225^{\circ},405^{\circ}$
    $\displaystyle x=22.5^{\circ},67.5^{\circ},157.5^{\circ}$ or $\displaystyle x=67.5^{\circ},112.5^{\circ},202.5^{\circ}$

    Ans.
    (From text book): $\displaystyle 22.5^{\circ},67.5^{\circ}$
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric Functions

    It is odd that the stated interval for $\displaystyle x$ is given in radians, yet the roots are given in degrees. Anyway, we are told:

    $\displaystyle 0\le x\le 180^{\circ}$ which means:

    $\displaystyle 0\le 2x\le 360^{\circ}$

    From the given equation, we find:

    $\displaystyle 2x=45^{\circ},\,135^{\circ}$ hence:

    $\displaystyle x=22.5^{\circ},\,67.5^{\circ}$
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  3. #3
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    Re: Trigonometric Functions

    Ok, thank you. My only other question then is why, based on the 2 answer outcome to the problem in the first post, is the answer to the similar question of $\displaystyle \cos3x=-\frac{1}{2}$ for $\displaystyle 0\leq x\leq180$ shown as a three solution answer$\displaystyle x=40,80,160$?

    Following, on from above I get:
    $\displaystyle \cos3x=\frac{1}{2}\rightarrow3x=60$. Cos is negative in the 2nd & 3rd quadrants.
    $\displaystyle 3x=180-60=120$ or $\displaystyle 3x=180+60=240$
    $\displaystyle x=40,80$
    How do I account for the 160?
    Last edited by GrigOrig99; Nov 12th 2012 at 10:54 AM.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric Functions

    With:

    $\displaystyle 0\le x\le 180^{\circ}$

    we have:

    $\displaystyle 0\le 3x\le 540^{\circ}$

    So, we find:

    $\displaystyle 3x=120^{\circ},\,240^{\circ},\,480^{\circ}$ and so:

    $\displaystyle x=40^{\circ},\,80^{\circ},\,160^{\circ}$
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  5. #5
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    Re: Trigonometric Functions

    Ok, thank you.
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