determining signs of trigonometric functionsDeterminingSignsOfTrigonometricFunctions2006-07-22 10:40:542007-05-25 01:19:43Topic\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{pstricks}There are at least two mnemonic devices for determining the sign \PMlinkescapetext{of} a trigonometric function at a given angle. They are ``all snow tastes cold'' and (the more mathematical version) ``all students take calculus''.
The first \PMlinkescapetext{word} in both of these, ``all'', indicates that, if an angle lies in the first quadrant, then, when any trigonometric function is applied to it, the result is positive.
The second \PMlinkescapetext{word} in both of these starts with the letter ``s'', which indicates that, if the terminal ray of an angle lies in the second quadrant, then the only trigonometric functions that can be applied to it that yield a positive result are $\sin$ and its reciprocal $\csc$.
The third \PMlinkescapetext{word} in both of these starts with the letter ``t'', which indicates that, if the terminal ray of an angle lies in the third quadrant, then the only trigonometric functions that can be applied to it that yield a positive result are $\tan$ and its reciprocal $\cot$.
The fourth \PMlinkescapetext{word} in both of these starts with the letter ``c'', which indicates that, if the terminal ray of an angle lies in the fourth quadrant, then the only trigonometric functions that can be applied to it that yield a positive result are $\cos$ and its reciprocal $\sec$.
Because of how these mnemonic devices work, it is clear that they are in \PMlinkescapetext{terms} of the calculator trigonometric functions.
Below is a picture that illustrates how the mnemonic device ``all students take calculus'' works:
\begin{center}
\begin{pspicture}(-3,-3)(3,3)
\psline{<->}(-3,0)(3,0)
\psline{<->}(0,-3)(0,3)
\rput[l](0.5,2.5){all are}
\rput[l](0.5,2){positive}
\rput[l](0.5,1.5){here}
\rput[l](0.3,0.3){I ``all''}
\rput[r](-0.5,2.5){$\sin$ and $\csc$}
\rput[r](-0.5,2){are positive}
\rput[r](-0.5,1.5){here}
\rput[r](-0.3,0.3){``students'' II}
\rput[r](-0.3,-0.3){``take'' III}
\rput[r](-0.5,-1.5){$\tan$ and $\cot$}
\rput[r](-0.5,-2){are positive}
\rput[r](-0.5,-2.5){here}
\rput[l](0.3,-0.3){IV ``calculus''}
\rput[l](0.5,-1.5){$\cos$ and $\sec$}
\rput[l](0.5,-2){are positive}
\rput[l](0.5,-2.5){here}
\rput[a](3,-0.2){$x$}
\rput[r](-0.2,3){$y$}
\rput[l](-3,0){.}
\rput[b](0,-3){.}
\end{pspicture}
\end{center}
For angles whose terminal ray lies on the boundary of two quadrants, the matter of determining sign is not as \PMlinkescapetext{simple}, but it is still possible to do so through use of the mnemonic device. If, for both of the boundary quadrants, the sign \PMlinkescapetext{of} the trigonometric function is positive, then the value of the trigonometric function applied to the angle is $1$. If, for both of the boundary quadrants, the sign \PMlinkescapetext{of} the trigonometric function is negative, then the value of the trigonometric function applied to the angle is $-1$. If the sign \PMlinkescapetext{of} the trigonometric function is different in the two boundary quadrants, then the value of the trigonometric function applied to the angle is either $0$ or undefined.
{\sl Example:\/} Since the terminal ray of $\displaystyle \frac{2\pi}{3}$ lies in the second quadrant, we have that $\displaystyle \sin \left( \frac{2\pi}{3} \right)>0$, and $\displaystyle \cos \left( \frac{2\pi}{3} \right)<0$.
$\bigg($ In fact, $\displaystyle \sin \left( \frac{2\pi}{3} \right)=\frac{\sqrt{3}}{2}$ and $\displaystyle \cos \left( \frac{2\pi}{3} \right)=\frac{-1}{2}$. $\bigg)$
{\sl Example:\/} Since the terminal ray of $\displaystyle \frac{7\pi}{2}$ lies on the boundary of the third and fourth quadrants and, when $\csc$ is applied to any angle whose terminal ray lies in either the third or fourth quadrant, the result is negative, we have that $\displaystyle \csc \left( \frac{7\pi}{2} \right)=-1$.