How can I convert cos(i) to standard rectangular form?

Question

What's the standard rectangular form of cos(i)?

Answer 1

I assume by i you mean a pure imaginary. Use the identity:

cos(z) = (e^(iz) + e^(-iz))/2

Substituting z = i: cos(i) = (e^(i^2) + e(-i^2))/2 = (e + 1/e)/2 = 1.543

Interestingly, the cosine of a pure inaginary is always real.

If you aren't familiar with the identity above, it is easiliy derivable from the Euler identity:

e^(iz) = cos(z) + i sin(z)

Answer 2

cos(i b) = 1/2(e^b + e^-b)
In other words, cos(i b) = cosh(b)

Therefore for b = 1,
cos(i) = 1/2 (e + 1/e)

Is that what you meant? This is derived from the Euler formulas:
cos(x) = 1/2(e^(i x) + e^(-i x))
sin(x) = 1/(2 i) (e^(i x) - e^(-i x))

Answer 3

Search out, and there are may good trig. illustrations, cos(i). Cos(i) to me represents the x value of a unit vector. If it is cos(theta), theta is the angle, then you have to assume a radius, r, thus cos(theta) r = x. and sin(theta) r = y. x, y is standard rectangular form.

Thus the rectangular coordinates, (x,y) are cos(theta) r, sin(theta) r.