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Trigonometric Functions and Complex Numbers


Date: 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
    
Associated Topics:
College Imaginary/Complex Numbers
College Trigonometry
High School Imaginary/Complex Numbers
High School Trigonometry

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