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Thread: When theta must be positive and when negative

  1. #1
    Junior Member B9766's Avatar
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    When theta must be positive and when negative

    I'm grappling with just not seeing when to use negative angles and when to use positive angles when solving trig functions.

    If I pick a point on a graph with coordinates (5,-5) the angle formed between the x-axis and the hypotenuse can be either $-45^\circ$ or $315^\circ$ (along with all multiples of $360^\circ$). Yes?

    If this is true, then I would think that $\sin(-45^\circ) = \sin(315^\circ) = \dfrac{-\sqrt{2}}{2}$

    If I plot the sine function I see that $\sin(-45^\circ) = \dfrac{-\sqrt{2}}{2}$ but my calculator gives me an error when entering $\sin(-45^\circ)$ or $\sin(\dfrac{-\pi}{4})$

    Likewise, my current lesson on complex numbers requires that I use the sin and cos functions of $\dfrac{7\pi}{4}$ rather than $\dfrac{-\pi}{4}$

    I would appreciate an explanation of when to use a negative reference angle and when to use the positive angle $< 360^\circ$
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    Re: When theta must be positive and when negative

    Quote Originally Posted by B9766 View Post
    I'm grappling with just not seeing when to use negative angles and when to use positive angles when solving trig functions.

    If I pick a point on a graph with coordinates (5,-5) the angle formed between the x-axis and the hypotenuse can be either $-45^\circ$ or $315^\circ$ (along with all multiples of $360^\circ$). Yes?
    If this is true, then I would think that $\sin(-45^\circ) = \sin(315^\circ) = \dfrac{-\sqrt{2}}{2}$
    If I plot the sine function I see that $\sin(-45^\circ) = \dfrac{-\sqrt{2}}{2}$ but my calculator gives me an error when entering $\sin(-45^\circ)$ or $\sin(\dfrac{-\pi}{4})$
    Likewise, my current lesson on complex numbers requires that I use the sin and cos functions of $\dfrac{7\pi}{4}$ rather than $\dfrac{-\pi}{4}$
    I would appreciate an explanation of when to use a negative reference angle and when to use the positive angle $< 360^\circ$
    Some older texts & authors still use degrees as well as reference angles which acute. If the terminal side is in quadrant II the reference angle is $180^{\circ}-\theta$;
    terminal side is in quadrant III the reference angle is $180^{\circ}+\theta$;terminal side is in quadrant IV the reference angle is $360^{\circ}-\theta$.
    In II the $\sin$ is positive, in III the $\tan$ is positive, & in IV the $cos$ is positive.

    Look Here. The basic parts of wolframalpha are free to use. The whole site is inexpensive for students & the is a version for small screen devices.
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    Re: When theta must be positive and when negative

    I don't know why your calculator can't determine sin(-pi/4)). Mine is ok with it, as well as sin(7 pi/4)

    Whether to measure angles as clockwise from the x-axis or counterclockwise depends on the application, and may be influenced by the particulars of the problem you are working. As an example, to calculate the y-coordinate of a point on a wheel that is rotating about the origin you would use y=R sin(theta) where theta is positive if the wheel rotates counterclockwise and theta is negative if it's rotating clockwise. This allows for a nice smooth plot of theta and y versus time. As for complex analysis - you are correct that by convention angles are measured clockwise from the real axis, and of course they are given in radians.
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    Junior Member B9766's Avatar
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    Re: When theta must be positive and when negative

    Really, thanks ChipB.

    As for the calculator, it was user error. Much to my chagrin, I keep forgetting the TI-84 requires the use of the (-) key rather than the minus key for negative values.

    My Trig and Pre-Calc course is a self-study program. I worry sometimes when I don't come up with the same answers as the text.

    As an example:

    Find Trigonometric form of $5 - 5i$. I came up with $z = 5\sqrt{2} (\cos\dfrac{-\pi}{4}+i \sin\dfrac{-\pi}{4})$

    But the text answer was $z = 5\sqrt{2} (\cos\dfrac{7\pi}{4}+i \sin\dfrac{7\pi}{4})$

    From your answer it seems either should be accepted as correct. Is that true? Would there normally be a preference?
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    Re: When theta must be positive and when negative

    Quote Originally Posted by B9766 View Post
    Really, thanks ChipB.
    As for the calculator, it was user error. Much to my chagrin, I keep forgetting the TI-84 requires the use of the (-) key rather than the minus key for negative values.
    My Trig and Pre-Calc course is a self-study program. I worry sometimes when I don't come up with the same answers as the text.
    As an example:
    Find Trigonometric form of $5 - 5i$. I came up with $z = 5\sqrt{2} (\cos\dfrac{-\pi}{4}+i \sin\dfrac{-\pi}{4})$
    But the text answer was $z = 5\sqrt{2} (\cos\dfrac{7\pi}{4}+i \sin\dfrac{7\pi}{4})$
    From your answer it seems either should be accepted as correct. Is that true? Would there normally be a preference?
    The "angle" associated with a complex is known as its argument. So $\displaystyle \arg (5 - 5\bf{i}) = - \frac{\pi }{4}$
    Note that I used the negative value. In resent times that has become the preferred notation. However older texts often used the convention that
    $\displaystyle 0\le \arg (z) <2\pi$ that seems to be the case with your textbook.
    Just note that $2\pi -\frac{\pi}{4}=\frac{7\pi}{4}$ That means that the two are equivalent.
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    Junior Member B9766's Avatar
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    Re: When theta must be positive and when negative

    Thanks for the confirmation, Plato.
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