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Version 7 |
| 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''. |
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''. |
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| 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 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. |
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| 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 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 $\csc$. |
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| 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 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 $\cot$. |
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| 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 $\sec$. |
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 $\sec$. |
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| Because of how these mnemonic devices work, it is clear that they are in \PMlinkescapetext{terms} of the calculator trigonometric functions. |
Because of how these mnemonic devices work, it is clear that they are in \PMlinkescapetext{terms} of the calculator trigonometric functions. |
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| Below is a picture that illustrates how the mnemonic device ``all students take calculus'' works: |
Below is a picture that illustrates how the mnemonic device ``all students take calculus'' works: |
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| \begin{center} |
\begin{center} |
| \begin{pspicture}(-3,-3)(3,3) |
\begin{pspicture}(-3,-3)(3,3) |
| \psline{<->}(-3,0)(3,0) |
\psline{<->}(-3,0)(3,0) |
| \psline{<->}(0,-3)(0,3) |
\psline{<->}(0,-3)(0,3) |
| \rput[l](0.5,2.5){all are} |
\rput[l](0.5,2.5){all are} |
| \rput[l](0.5,2){positive} |
\rput[l](0.5,2){positive} |
| \rput[l](0.5,1.5){here} |
\rput[l](0.5,1.5){here} |
| \rput[l](0.3,0.3){I ``all''} |
\rput[l](0.3,0.3){I ``all''} |
| \rput[r](-0.5,2.5){$\sin$ and $\csc$} |
\rput[r](-0.5,2.5){$\sin$ and $\csc$} |
| \rput[r](-0.5,2){are positive} |
\rput[r](-0.5,2){are positive} |
| \rput[r](-0.5,1.5){here} |
\rput[r](-0.5,1.5){here} |
| \rput[r](-0.3,0.3){``students'' II} |
\rput[r](-0.3,0.3){``students'' II} |
| \rput[r](-0.3,-0.3){``take'' III} |
\rput[r](-0.3,-0.3){``take'' III} |
| \rput[r](-0.5,-1.5){$\tan$ and $\cot$} |
\rput[r](-0.5,-1.5){$\tan$ and $\cot$} |
| \rput[r](-0.5,-2){are positive} |
\rput[r](-0.5,-2){are positive} |
| \rput[r](-0.5,-2.5){here} |
\rput[r](-0.5,-2.5){here} |
| \rput[l](0.3,-0.3){IV ``calculus''} |
\rput[l](0.3,-0.3){IV ``calculus''} |
| \rput[l](0.5,-1.5){$\cos$ and $\sec$} |
\rput[l](0.5,-1.5){$\cos$ and $\sec$} |
| \rput[l](0.5,-2){are positive} |
\rput[l](0.5,-2){are positive} |
| \rput[l](0.5,-2.5){here} |
\rput[l](0.5,-2.5){here} |
| \rput[a](3,-0.2){$x$} |
\rput[a](3,-0.2){$x$} |
| \rput[r](-0.2,3){$y$} |
\rput[r](-0.2,3){$y$} |
| \rput[l](-3,0){.} |
\rput[l](-3,0){.} |
| \rput[b](0,-3){.} |
\rput[b](0,-3){.} |
| \end{pspicture} |
\end{pspicture} |
| \end{center} |
\end{center} |
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| 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. |
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. |
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| {\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 $\displaystyle \cos \left( \frac{2\pi}{3} \right)<0$. |
{\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 $\displaystyle \cos \left( \frac{2\pi}{3} \right)<0$. |
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| $\bigg($ In fact, $\displaystyle \sin \left( \frac{2\pi}{3} \right)=\frac{\sqrt{3}}{2}$ and $\displaystyle \cos \left( \frac{2\pi}{3} \right)=\frac{-1}{2}$. $\bigg)$ |
$\bigg($ In fact, $\displaystyle \sin \left( \frac{2\pi}{3} \right)=\frac{\sqrt{3}}{2}$ and $\displaystyle \cos \left( \frac{2\pi}{3} \right)=\frac{-1}{2}$. $\bigg)$ |
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| {\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 $\displaystyle \csc \left( \frac{7\pi}{2} \right)=-1$. |
{\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 $\displaystyle \csc \left( \frac{7\pi}{2} \right)=-1$. |
