Question 1122398
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It is called the Pythagorean Theorem.  If *[tex \Large (x, y)] is on the terminal ray of a triangle, then *[tex \Large \sqrt{x^2\,+\,y^2}] is the hypotenuse of the right triangle formed by the line segment from the origin to the point, the vertical line segment from the point to the *[tex \Large x]-axis, and the portion of the *[tex \Large x]-axis from the origin to the the point *[tex \Large (x,0)].  The hypotenuse, from the definition of the sin and cos functions, is the denominator of the fractions that represent the sin and cos ratios.  Then *[tex \Large x] is the numerator of the cos, and *[tex \Large y] is the numerator of sin.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \frac{y}{\sqrt{x^2\,+\,y^2}}]


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


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


Then cot is the reciprocal of tan, sec is the reciprocal of cos, and csc is the reciprocal of sin.


Just plug in your numbers and do the arithmetic.  I presume you know how to rationalize a denominator.  If not, there are a couple of lessons on this site to teach that.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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