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Thread: [SOLVED] description question about reference angles.

  1. #1
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    [SOLVED] description question about reference angles.

    Explain how the reference angle is used to find values of the trigonometric functions for an angle in quadrant III.

    ?? i don't know.
    and also, why is the reference angle always found with reference to the x-axis and not the y-axis?
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  2. #2
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    Hello somanyquestions
    Quote Originally Posted by somanyquestions View Post
    Explain how the reference angle is used to find values of the trigonometric functions for an angle in quadrant III.

    ?? i don't know.
    and also, why is the reference angle always found with reference to the x-axis and not the y-axis?
    The reference angle is always measured anti-clockwise from the positive direction of the $\displaystyle x$-axis.

    Your question about why not the $\displaystyle y$-axis is pointless. A definition is a definition. You may define a different reference angle starting from the $\displaystyle y$-axis if you wish - or indeed any other line - but in doing so, you will be creating a different definition. You might as well ask "Why doesn't the alphabet start at the letter 'n'?". Without some agreed conventions, we should get nowhere.

    OK then. If the point $\displaystyle (x, y)$ is on the unit circle, and the radius joining $\displaystyle (0, 0)$ to $\displaystyle (x,y)$ makes an angle $\displaystyle \theta$ with the positive direction of the $\displaystyle x$-axis, measured anticlockwise, as described above, then, by definition:
    $\displaystyle \left\{\begin{array}{l}
    \cos\theta = x\\
    \sin\theta = y\\
    \end{array}\right.$
    This definition holds good for all values of $\displaystyle \theta$; i.e. for angles in all quadrants. In particular, in QIII, where $\displaystyle x<0$ and $\displaystyle y<0$, this will mean that $\displaystyle \cos\theta$ and $\displaystyle \sin\theta$ are both negative.

    Draw a diagram, showing an angle $\displaystyle \theta$ between $\displaystyle \pi$ and $\displaystyle 3\pi/2$ (i.e. $\displaystyle (x,y)$ lies in QIII). Now draw the diameter through $\displaystyle (x,y)$ to meet the circle again in QI. You should be able to see that this shows that:
    $\displaystyle \left\{\begin{array}{l}
    \cos\theta = - \cos(\theta-\pi)\\
    \sin\theta = -\sin(\theta-\pi)\\
    \end{array}\right.$
    where $\displaystyle 0 < (\theta - \pi) < \pi/2$; i.e. $\displaystyle (\theta - \pi)$ is an acute angle. This shows how to relate the values of sine and cosine in QIII to those in QI.

    Grandad
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