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Thread: principle angles and related acute angles

  1. #1
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    principle angles and related acute angles

    how do you find them?
    examples:
    tan18
    sin205
    cos-55
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  2. #2
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    Re: principle angles and related acute angles

    Quote Originally Posted by ahmedb View Post
    how do you find them?
    examples:
    tan18
    sin205
    cos-55
    I'm assuming these are in degrees.

    $\displaystyle \displaystyle \begin{align*} \sin{(5x)} &= 16\sin^5{(x)} - 20\sin^3{(x)} + 5\sin{(x)} \\ \sin{(5\cdot 18^{\circ})} &= 16\sin^5{(18^{\circ})} - 20\sin^3{(18^{\circ})} + 5\sin{(18^{\circ})} \\ 1 &= 16\sin^5{(18^{\circ})} - 20\sin^3{(18^{\circ})} + 5\sin{(18^{\circ})} \\ 0 &= 16\sin^5{(18^{\circ})} - 20\sin^3{(18^{\circ})} + 5\sin{(18^{\circ})} - 1 \\ 0 &= 16y^5 - 20y^3 + 5y - 1 \textrm{ if we let }y = \sin{(18^{\circ})} \\ 0 &= (y - 1)\left(4y^2 + 2y - 1\right)^2 \\ 0 &= 4y^2 + 2y - 1 \\ 0 &= y^2 + \frac{1}{2}y - \frac{1}{4} \\ 0 &= y^2 + \frac{1}{2}y + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{4} \\ 0 &= \left( y + \frac{1}{4}\right)^2 - \frac{5}{16} \\ \frac{5}{16} &= \left(y + \frac{1}{4}\right)^2 \\ \pm \frac{\sqrt{5}}{4} &= y + \frac{1}{4} \\ \frac{-1 \pm \sqrt{5}}{4} &= y \end{align*} $

    With a little checking, you'll see that $\displaystyle \displaystyle \begin{align*} \sin{(18^{\circ})} = \frac{-1 + \sqrt{5}}{4} \end{align*} $ is the correct solution.

    Now use a similar method using

    $\displaystyle \displaystyle \begin{align*} \cos{(5x)} = 16\cos^5{(x)} - 20\cos^3{(x)} + 5\cos{(x)} \end{align*} $ to get the value of $\displaystyle \displaystyle \begin{align*} \cos{(18^{\circ})} \end{align*} $.

    Then you can find $\displaystyle \displaystyle \begin{align*} \tan{(18^{\circ})} \end{align*} $ using $\displaystyle \displaystyle \begin{align*} \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}} \end{align*} $.
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  3. #3
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    Re: principle angles and related acute angles

    what i mean is like quadrant 1, 2, 3, and 4.
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  4. #4
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    Re: principle angles and related acute angles

    Here's a little easier method:

    $\displaystyle \begin{align*} x &= 18^{\circ} \\ 5x &= 90^{\circ} \\ 2x+ 3x &= 90^{\circ} \\ 2x &= 90^{\circ} -3x \\ \sin(2x)&= \sin(90^{\circ}-3x) \\ \sin(2x) &= \cos(3x) \\ 2\sin(x)\cos(x) &= \cos^3(x)-3\sin^2(x)\cos(x) \\ 2\sin(x) &= \cos^2(x)-3\sin^2(x) \\ 2\sin(x) &= 1-\sin^2(x)-3\sin^2(x) \\ 4\sin^2(x)+2\sin(x)-1 &= 0 \\ \sin(x) &= \frac{-1 \pm \sqrt{5}}{4} \quad [\text{Quadratic Formula}] \\ \sin(18^{\circ}) &= \frac{-1 + \sqrt{5}}{4} \end{align*}$


    $\displaystyle \sin(205 ^{\circ})=-\sin(25 ^{\circ} )$

    $\displaystyle \cos(-55 ^{\circ}) = \cos(55 ^{\circ}) $

    Unfortunately, $\displaystyle \sin(25 ^{\circ} )$ and $\displaystyle \cos(55 ^{\circ})$ have no exact values.
    Last edited by sbhatnagar; Dec 19th 2011 at 05:26 AM.
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  5. #5
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    Re: principle angles and related acute angles

    Quote Originally Posted by ahmedb View Post
    what i mean is like quadrant 1, 2, 3, and 4.
    ... are you asking about reference angles?
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