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 it then follows that b = arccosh(4)
Thanks for getting back so promptly , unfortunately i do not what you mean by take sinus of both sides. I have got the stage you have it is the very final stage I van not do
I apologize for the confusion. Let me explain step by step how to get the expression for arcsin(4) in the form a+ib.
1. Start with the equation: 4 = sin(a)cosh(b) + i cos(a)sinh(b)
2. To simplify, we need to take the inverse hyperbolic cosine (arccosh) of both sides. This will isolate the b term.
3. Apply the arccosh function to both sides of the equation. The arccosh function is the inverse of the hyperbolic cosine function.
4. By applying arccosh to both sides, we get: arccosh(4) = arccosh(sin(a)cosh(b) + i cos(a)sinh(b))
5. Now, to determine the value of a, we can look at the real part of the equation. The real part is sin(a)cosh(b). If we equate this to zero, we can solve for a.
6. Setting sin(a)cosh(b) = 0 gives us sin(a) = 0. This means that a must be pi/2 (90 degrees) or any integer multiple of pi/2, as sin(a) = 0 for those values of a.
7. With a = pi/2, the equation simplifies to: arccosh(4) = arccosh(cosh(b))
8. Now, we can solve for b by isolating it on one side. This gives us b = arccosh(4).
So, in summary, arcsin(4) in the form a+ib is a = pi/2 and b = arccosh(4).