Question 1169026
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The three points *[tex \Large (0,0)], *[tex \Large (3,0)], and *[tex \Large (3,1)] form a right triangle with angle *[tex \Large \theta] at the origin.


First calculate the measure of the hypotenuse, which is the distance from the origin to point *[tex \Large (3,1)].  Hint: use Pythagoras.


The sine of *[tex \Large \theta] is the opposite side (the measure of the segment with endpoints *[tex \Large (3,0)] and *[tex \Large (3,1)]) divided by the measure of the hypotenuse that you just calculated.  (Don't forget to rationalize your denominator)


The cosine of *[tex \Large \theta] is the adjacent side (the measure of the segment with endpoints *[tex \Large (0,0)] and *[tex \Large (3,0)]) divided by the hypotenuse.


Once you know the sine and cosine of any angle, you can calculate the other four functions thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(\theta)\ =\ \frac{\sin(\theta)}{\cos(\theta)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot(\theta)\ =\ \frac{1}{\tan(\theta)}\ =\ \frac{\cos(\theta)}{\sin(\theta)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec(\theta)\ =\ \frac{1}{\cos(\theta)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc(\theta)\ =\ \frac{1}{\sin(\theta)]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 




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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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