Solving Trig Equations with Newton's Method
Date: 06/16/2006 at 01:05:31 From: Sam Subject: Solving trigonometry equations What is the technique to solve a trigonometric equation like this? Do I need to use calculus? I've studied a bit of that. 3sin(x) = x + 1 (0 < x < 2pi) What if the above equation is in quadratic or some higher order, what is the technique to do that?
Date: 06/16/2006 at 09:43:36 From: Doctor Jerry Subject: Re: Solving trigonometry equations Hello Sam, Thanks for writing to Dr. Math. This equation is of "higher order;" it is called a "transcendental equation." The needed technique is either numerical or graphical. I'll assume that you can do the graphical with the help of a calculator. Actually, most "scientific" calculators these days can also solve this kind of equation numerically. If you want to do it "by hand," you can use something like Newton's Method, which uses calculus. Take a look at this figure: Notice that I've graphed in the left figure the two original graphs. In the right figure I've graphed the function f(x) = 3sin(x) - (x+1). We want to find its two zeros. I'll concentrate on the left one, near 0.53. In Newton's Method, one makes a guess as to where the root is. In this case, I'll guess that the root is near 0.75. Call this x1. We work out a sequence of guesses, which usually improve and converge to the root. To find x2 we do this: From (x1,0) we go vertically until we hit the graph of f. At that point we draw a tangent line. Call the x-intercept of this tangent line x2. Once we have x2, we repeat this procedure. It's not difficult to show that x2 = x1 - f(x1)/f'(x1) x3 = x2 - f(x2)/f'(x2) and so on. If x1 = 0.75, then x2 = 0.503222 x3 = 0.537907 x4 = 0.53847 We may be reasonably confident that the root is near 0.538. We can check this by calculating f(0.538). We find f(0.538) = -0.000741359. We can continue generating new approximations until we achieve the accuracy we want. Feel free to write back if my comments are not clear or you need more help on this problem. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/
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