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There are at least two mnemonic devices for determining the sign 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 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 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 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 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 terms of the calculator trigonometric functions.
Below is a picture that illustrates how the mnemonic device ``all students take calculus'' works:
\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}
For angles whose terminal ray lies on the boundary of two quadrants, the matter of determining sign is not as simple, but it is still possible to do so through use of the mnemonic device. If, for both of the boundary quadrants, the sign 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 of the trigonometric function is negative, then the value of the trigonometric function applied to the angle is $-1$ . If the sign 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.
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 $\displaystyle \cos \left( \frac{2\pi}{3} \right)=\frac{-1}{2}$ . $\bigg)$
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 $\displaystyle \csc \left( \frac{7\pi}{2} \right)=-1$ .
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