# Maths

posted by .

use the formula
cos(x+iy)=cosxcosiy-sinxsiniy
to find two imaginary numbers whose cosine is 3

cos(x+iy)=cosxcosiy-sinxsiniy =

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

gives:

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

Thanks

thanks

## Similar Questions

1. ### trig

Reduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The quantity …
2. ### tigonometry

expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b) …
3. ### Trig

Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v - u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos …
4. ### Maths

How do I write arcsin4 in the form a+ib. I know the answer but do not Know how to get to it Take sinus of both sides: 4 = sin(a)cosh(b) + i cos(a)sinh(b) Equate real and imaginary parts. You then see that you can put a = pi/2. And …
5. ### maths 2

The original question I had was write arcsin4 in the form a+ib. I manage and understand how to get so far BUT How do I get from cosacoshb-isinasinhb=4 to 2m(pi)+/- iarccosh4 arcsin4 = a + b i ---> 4 = sin(a + bi) sin(a + bi) = sin(a)cos(bi) …

7. ### Calculus II

What is sin(i) -- (sin of the imaginary number, i) What is cos(i)-- cosine of the imaginary number, i orrr tan(1+i) You can derive these things from the equation: Exp(i x) = cos(x) + i sin(x) ---> sin(x) = [Exp(ix) - Exp(-ix)]/(2i) …
8. ### Mathematics - Trigonometric Identities

Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y …
9. ### TRIG!

Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 …