Studying Mathematics Thread

[polymath257]

Alternatively, it is the cosine and sine sum formulas applied repeatedly by induction.

But wait... seriously... why do you just get to put the exponent in front of the cos and isin? How does that work? See, I once had this professor who insisted on proving to us why a formula worked, and I'd sit there in class thinking, "I don't give a shit! This is boring and confusing. Just give me the formula and I'll plug in my little values and get an A in your class. Cause that's what I do!"

But now I'm in this place where I'm seeing things and I can't just take DeMovire's word for it. I'll grant that I'm more inclined to take Euler's word for it, but I'd still like to know why this formula works.

Well, first, do you know the sine and cosine sum formulas?

sin(A+B)=sin(A)cos(B)+cos(A)sin(B)
cos(A+B)=cos(A)cos(B)-sin(A)sin(B)

If not, then we can go through those separately. If so, then see what happens when you multiply out

[cos(A)+i*sin(A)] * [cos(B)+i*sin(B)]

You should get

cos(A)cos(B) + i* sin(A)cos(B) +i*cos(A)sin(B) +i^2 *sin(A)sin(B)

=[cos(A)cos(B)-sin(A)sin(B)]+i*[sin(A)cos(B)+cos(A)sin(B)] =cos(A+B) +i *sin(A+B)

(remember that i^2 =-1)

In other words, if you multiply two expressions of the form cos(A)+i*sin(A), the result can be obtained by *adding* the angles involved.

Now, what happens if you multiply the *same* expression over and over again? The angle adds up again and again, however many times you did the multiplication.

That is why

[cos(A)+i*sin(A)]^n = cos(n A)+i*sin(n A)

Each multiplication corresponds to an addition of the angles.