Question

Precalc help please :)

Answer 1

have you considered a more general approach?

Answer 2

n (cos(a1) + i sin(a1) )
m (cos(b1) + i sin(b1) )
-------------------
nm [cos(a1)cos(b1) +i^2 sin(a1)sin(b1) + i cos(a1)sin(b1) + i cos(b1)sin(a1)]

i^2 = -1 soo

nm [cos(a1)cos(b1) - sin(a1)cos(b1) + i (cos(a1)sin(b1) + cos(b1)sin(a1))]

what does this simplify to when we use trig identities?
---------------------

n (cos(a1) + i sin(a1) )
-----divide by------
m (cos(b1) + i sin(b1) )

n/m is fine, we want to conjugate the bottom to form a real number

cos(a1) + i sin(a1) cos(b1) - i sin(b1)
----divide by---- ----divide by----
cos(b1) + i sin(b1) cos(b1) - i sin(b1)

but -sin(t) = sin(-t), and cos(t) = cos(-t)


[cos(a1) + i sin(a1)] [cos(-b1) + i sin(-b1)]
--------------divide by--------------
cos^2(b1) + sin^2(b1)

the bottom is 1 by identity, and the top is just multiplying again

Answer 3

a concise form for this is simply written out as

(n,a) (m,b) = (nm,a+b)

Answer 4

(n,a)/(m/b) = (n/m, a-b)

Answer 5

hmm wht does that look like with my numbers? o.0

Answer 6

Oo! I understand now :)
thanks for all your hard work in writing it all out!

Answer 7

:) the work was more for my remembering, since i dont get to work these often i tend to forget the formulas, and have to reinvent them all over again