Question 972990
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Trick question.  Except for cotangent, you cannot find THE value of the other functions, you can only find all possible values of the other functions.


For example, let *[tex \Large \tan\theta\ =\ x]


Then from the definition of tangent,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin\theta}{\cos\theta}\ =\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ x\cos\theta]


Square both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ x^2\cos^2\theta]


Use the Pythagorean Identity


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2\theta\ =\ x^2\cos^2\theta]


Then do a bunch of algebra that I will let you verify on your own:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \pm\sqrt{\frac{1}{x^2\ +\ 1}}]


Similarly


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ x\ \Rightarrow\ \sin\theta\ =\ \pm\sqrt{\frac{x^2}{x^2\ +\ 1}]


And the other three functions are just the reciprocals of the ones we now know.


However, for any given non-zero value of *[tex \Large \tan\theta], there are two values for sine, cosine, secant, and cosecant.  So, depending on exactly what your instructor meant by <i>"...find the values of all the other trigonometric functions without finding the actual measure of the angle"</i>, the answer to your question is either yes or no.


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

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