Proofs

$e^{i\pi }+1=0$ (Euler's identity)

$e^{i\theta }=\cos \theta +i\sin \theta$ (Euler's formula)

$(\cos \theta +i\sin \theta {)}^n=\cos n\theta +i\sin n\theta$ (de Moivre's formula)

$(a+b{)}^2=a^2+2ab+b^2$

$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$

$(a+b{)}^n={\displaystyle \sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k}$ (Binomial theorem)

$a^2-b^2=(a+b)(a-b)$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

$a^n-b^n={\displaystyle (a-b)\sum _{k=0}^{n-1}{a}^{n-k-1}{b}^k}$

$a{x}^2+bx+c=0$

$⟹x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$a^m{a}^n=a^{m+n}$

$(a^m{)}^n=a^{mn}$

$(ab{)}^n=a^n{b}^n$

${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$

${\log }_b(x^k)=k{\log }_{b}x$

${\log }_{b}x={\displaystyle \frac{{\log }_{c}x}{{\log }_{c}b}}$

${\sin }^2\theta +{\cos }^2\theta =1$ (Pythagorean identity)

$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$

$\sin 2\theta =2\sin \theta \cos \theta$

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =1-2{\sin }^2\theta =2{\cos }^2\theta -1$

$\sin n\theta ={\displaystyle \sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }(-1{)}^k(\genfrac{}{}{0ex}{}{n}{2k+1}){\cos }^{n-(2k+1)}\theta {\sin }^{2k+1}\theta }$

$\cos n\theta ={\displaystyle \sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }(-1{)}^k(\genfrac{}{}{0ex}{}{n}{2k}){\cos }^{n-2k}\theta {\sin }^{2k}\theta }$