User:Ybab321/Sandbox

Feel free to help or even complete whatever proofs appear here.

= Cosine Multiple Angle Formula =

Theorem

 * $\displaystyle \cos n \theta = n \sum^{\left\lfloor{\frac n 2}\right\rfloor}_{k \mathop = 0} \frac{\left({-1}\right)^k \left({n-k-1}\right)! 2^{n-2k-1} \cos^{n-2k} \theta}{k!\left({n-2k}\right)!}$

Proof 2
<!-- = Polynomial Expansion =

Theorem
Let
 * $f(n,k) = \begin{cases}

\{(\{\}, \bigcup^n_{i=1} \{i\})\} & : k = 0\\ \displaystyle \bigcup_{(a,b) \in f(n,k-1)} \bigcup^n_{i=\sup(a)+1} \{(a \cup \{i\},\ b \setminus \{i\})\} & : k \ne 0 \end{cases}$

Then:
 * $\displaystyle \prod^n_{k=1} (a_k x - b_k) =

\sum^n_{k=0} (-1)^{n-k} x^k \sum_{(p,q) \in f(n,k)} \prod_{i \in p} a_i \prod_{i \in q} b_i $

Proof by Mathematical Induction
Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \prod^n_{k=1} (a_k x - b_k) = \sum^n_{k=0} (-1)^{n-k} x^k \sum_{(p,q) \in f(n,k)} \prod_{i \in p} a_i \prod_{i \in q} b_i$

Basis for the Induction
$P \left({1}\right)$ is true, as this just says:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({n}\right)$ is true, where $n \ge 1$, then it logically follows that $P \left({n+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \prod^n_{k=1} (a_k x - b_k) = \sum^n_{k=0} (-1)^{n-k} x^k \sum_{(p,q) \in f(n,k)} \prod_{i \in p} a_i \prod_{i \in q} b_i$

Then we need to show:
 * $\displaystyle \prod^{n+1}_{k=1} (a_k x - b_k) = \sum^{n+1}_{k=0} (-1)^{n+1-k} x^k \sum_{(p,q) \in f(n+1,k)} \prod_{i \in p} a_i \prod_{i \in q} b_i$

Induction Step
This is our induction step:

Note that

So $P \left({n}\right) \implies P \left({n+1}\right)$ and the result follows by the Principle of Mathematical Induction. -->