Finding a Trig Function of an AngleDate: 07/09/2004 at 05:23:09 From: Sara Subject: trigonometry I need a short way to find the answer to this kind of question: If sin(y) = 8/17, find cot(y). I know cot(y) is cos(y)/sin(y), so cot(y) = (?/17)/(8/17) and cot(y) = ?/8 Date: 07/10/2004 at 11:34:10 From: Doctor Willae Subject: Re: trigonometry Dear Sara, There is a whole family of problems like these, and yours is the most basic. This is really a question about inverse trig functions. I'll use the prefix "arc" to mean an inverse trig function. Like this: arcsin(sin(x)) = x arcsin(1) = pi/2 because sin(pi/2) = 1 Going back to your original question, if the sin(y) = 8/17, we can determine y by just applying the inverse sine function: y = arcsin(sin(y)) = arcsin(8/17) Once we know angle y, we can determine the cotangent of y by just taking the tangent of y and inverting it: cot(y) = 1 / tan(y) = 1 / tan(arcsin(8/17)) Now...before you break out your calculator, you should really learn how to do this by hand. Fortunately, there's a trick that works for all of these problems. Draw yourself a right triangle. (I'll put one here, but you'll have to forgive my artwork.) Our goal will be to label all 3 sides of the triangle. Start by labeling one of the non-right-angles y. I'll put it in the bottom right. ** * ** * *** * *** * *** * *** * *** * y ** ********************** What do we know about the angle y? We know its sine: sin(y) = 8/17. And remember that sine is opposite over hypotenuse. So we can label two of the sides of our triangle: ** * ** * *** 17 * *** 8 * *** * *** * *** * y ** ********************** What else do we know? Pythagoras. If A and B are legs of a right triangle, and C is its hypotenuse, then: C^2 = A^2 + B^2. We've got the length of one leg and the hypotenuse. So we can solve for the other leg: B^2 = C^2 - A^2 B = sqrt(C^2 - A^2) = sqrt(17^2 - 8^2) = sqrt(289 - 64) = 15 Now we can finish labeling the sides of the triangle: ** * ** * *** 17 * *** 8 * *** * *** * *** * y ** ********************** 15 We're all but finished. You can read off the value of any trig function involving y from this triangle. You're interested in the cotangent, so let's look at that: cot(y) = adjacent/opposite = 15/8 But you could just as easily use this same technique to get the cosine of y: cos(y) = adjacent/hypotenuse = 15/17 Remember that at the beginning I said there was a whole family of this type of question. Yours is the most basic since we aren't told anything else about y, and therefore we just assume that y is an acute angle and draw the triangle the way we did. Sometimes in this type of problem, more information is given. Suppose we were also told that tan(y) < 0, or in other words the tangent of y is negative. That would tell us that y can't be acute, since all six trig functions of an acute angle are positive. But by understanding the use of reference angles and the signs of the trig functions by quadrant, we could determine that if sin(y) = 8/17 and tan(y) < 0, then y must be an angle in the second quadrant. We could then still draw our triangle, but we'd have to construct it in the second quadrant and the 8 would become -8. I don't want to go into more detail on that here since your question didn't involve it. I just wanted to mention that there are variations of this sort of problem where you must give consideration as to where to draw the triangle. But you can still use the triangle! This is a really useful trick to know, so let me know if anything isn't clear. I actually just used this same procedure to speed up some software I'm working on. Let me know how it goes. - Doctor Willae, The Math Forum http://mathforum.org/dr.math/ |
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