Posted by Tom on Sunday, January 21, 2007 at 5:08am.
use the formula
to find two imaginary numbers whose cosine is 3
cos(x)cosh(y) - i sin(x)sinh(y) = 3
Equating imaginary parts gives:
sin(x)sinh(y) = 0
You know that y cannot be zero, otherwise the complex number woyuld be real, but for real arguments the cosine is always between -1 and 1. If y is not zero, sinh(y) is not zero, and therefore sin(x) must be zero. This means that x is an integer times pi.
x = n pi
Equating the real parts gives:
cos(x)cosh(y) = 3
Insert x = n pi, using
cos(n pi) = (-1)^n
cosh(y) = 3*(-1)^n
cosh for real arguments is always positive, so n must be even. And we see that:
y = arccosh(3)
So, the complex numbers are of the form:
2n pi + arccosh(3)i
Answer This Question
More Related Questions
- maths 2 - The original question I had was write arcsin4 in the form a+ib. I ...
- Calculus 12th grade (double check my work please) - 1.)Find dy/dx when y= Ln (...
- tigonometry - expres the following as sums and differences of sines or cosines ...
- trig - Reduce the following to the sine or cosine of one angle: (i) sin145*cos75...
- Trig - Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < ...
- TRIG! - Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6...
- algebra - Can someone please help me do this problem? That would be great! ...
- Math - Calculus - The identity below is significant because it relates 3 ...
- Mathematics - Trigonometric Identities - Let y represent theta Prove: 1 + 1/tan^...
- Calculus II - What is sin(i) -- (sin of the imaginary number, i) What is cos(i...