For those of you who tried to derive the binomial expansion formula yourselves, here’s the step-by-step process below, based on how I derived the result when first studying complex numbers last year:
Setting \( x=\cos(\theta), y=i\sin(\theta) \) we can expand below:
(\cos(\theta)+i\sin(\theta))^n=\sum_{k=0}^{n}\binom{n}{k}\cos^{n-k}(\theta)(i\sin(\theta))^kBreaking apart the sum into real and imaginary terms by moving all the even terms to one sum and odd terms to another we get:
=\sum_{k=0}^{n/2}\binom{n}{2k}\cos^{n-2k}(\theta)(i\sin(\theta))^{2k} +\sum_{k=0}^{(n-1)/2}\binom{n}{2k+1}\cos^{n-2k-1}(\theta)(i\sin(\theta))^{2k+1}Finally, we can pull i out of the right-side sum and write \( i^{2k}=(\sqrt{-1})^k \) to get our final formula:
=\sum_{k=0}^{n/2}(-1)^k\binom{n}{2k}\cos^{n-2k}(\theta)\sin(\theta)^{2k} +i\sum_{k=0}^{(n-1)/2}(-1)^k\binom{n}{2k+1}\cos^{n-2k-1}(\theta)\sin(\theta)^{2k+1} We can test this for a couple of values of n and prove that they work: For example, n = 2 yields the familiar double-angle formulas for sine and cosine:
\sum_{k=0}^{2}(-1)^k\binom{2}{2k}\cos^{2-2k}(\theta)\sin(\theta)^{2k} +i\sum_{k=0}^{2}(-1)^k\binom{2}{2k+1}\cos^{2-2k-1}(\theta)\sin(\theta)^{2k+1}Hope this was helpful!
(x+y)^n=\binom{n}{0}x^n+\binom{n}{1}x^{n-1}y+\binom{n}{2}x^{n-2}y^2+\ldots=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k
The Binomial Theorem with Pascal’s Triangle.
Credit: [Mathematics Libretexts]