Trigonometric Functions and Complex NumbersDate: 27 Jun 1995 19:14:04 -0400 From: Anonymous Subject: Math Question Is there a solution to the following equation? Sin(a) = 5, where a = x + iy (complex value) Thanks for your help! Date: 28 Jun 1995 09:30:40 -0400 From: Dr. Ken Subject: Re: Math Question Hello there! To solve this question, you have to know a couple of facts about the Trigonometric functions. The first one is that sin(a + b) = sin(a)cos(b) + cos(a)sin(b) So in our problem, we have sin(a) = sin(x + iy) = sin(x)cos(iy) + cos(x)sin(iy) Now we're going to deal with those i's. We'll use the facts that sin(iy) = i*sinh(y) and cos(iy) = cosh(y). So we have sin(a) = sin(x + iy) = sin(x)cos(iy) + cos(x)sin(iy) = sin(x)cosh(y) + i*cos(x)sinh(y) So to get a solution for sin(a) = 5, we need to make that second part zero, and the first part 5. To do that, choose x=Pi and y=ArcCosh(5), which is about 2.29243. Thanks for the question! -K |
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