< Math formulas

Let $e^{ix}=f+ig$. (Eq. 1)

Differentiate both sides with respect to $x$:

$(e^{ix}{)}^{\prime }=(f+ig{)}^{\prime }$

$i{e}^{ix}=f^{\prime }+i{g}^{\prime }$ (Eq. 2)

Substitute Eq. 1 into the left-hand side:

$i(f+ig)=f^{\prime }+i{g}^{\prime }$

$-g+if=f^{\prime }+i{g}^{\prime }$

Equate the real and imaginary parts of both sides:

$-g=f^{\prime }$
$f=g^{\prime }$

Which can be rewritten as:

$f^{″}=-f$
$g^{″}=-g$

These are second-order differential equations. By substituting $x=0$ into Eqs. 1 and 2, we obtain the initial conditions $f(0)=1$, $f^{\prime }(0)=0$, $g(0)=0$, and $g^{\prime }(0)=1$. The unique solutions are:

$f=\cos x$
$g=\sin x$

Let $a$ and $b$ be real numbers, and let $m$ and $n$ be positive integers.

Definition (Exponentiation):

$a^1=a$, (Def. 1)

$a^{n+1}=a^{n}a$ for $n\ge 1$. (Def. 2)

By induction on $n$:

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

Base case ($n=1$):

$a^m{a}^1=a^{m}a$ (by Def. 1)

$=a^{m+1}$. (by Def. 2)

Inductive step: Assume $a^m{a}^n=a^{m+n}$. Then:

$a^m{a}^{n+1}=a^m(a^{n}a)$ (by Def. 2)

$=(a^m{a}^n)a$ (by associativity)

$=a^{m+n}a$ (by the induction hypothesis)

$=a^{(m+n)+1}$ (by Def. 2)

$=a^{m+(n+1)}$. (by associativity)

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

Base case ($n=1$):

$(a^m{)}^1=a^m$ (by Def. 1)

$=a^{m\cdot 1}$. (by identity)

Inductive step: Assume $(a^m{)}^n=a^{mn}$. Then:

$(a^m{)}^{n+1}=(a^m{)}^n{a}^m$ (by Def. 2)

$=a^{mn}{a}^m$ (by the induction hypothesis)

$=a^{mn+m}$ (by Rule 1)

$=a^{m(n+1)}$. (by distributivity)

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

Base case ($n=1$):

$(ab{)}^1=ab$ (by Def. 1)

$=a^1{b}^1$. (by Def. 1)

Inductive step: Assume $(ab{)}^n=a^n{b}^n$. Then:

$(ab{)}^{n+1}=(ab{)}^n(ab)$ (by Def. 2)

$=a^n{b}^n(ab)$ (by the induction hypothesis)

$=(a^{n}a)(b^{n}b)$ (by associativity and commutativity)

$=a^{n+1}{b}^{n+1}$. (by Def. 2)

Let $x,y>0$ and $k$ be real numbers, and let $b,c>0$ be real numbers with $b,c\ne 1$. Since a logarithm is the inverse of exponentiation, we have:

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

just like $f(f^{-1}(x))=f^{-1}(f(x))=x$. Using this:

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

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

$={\log }_{b}x+{\log }_{b}y$.

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

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

$=k{\log }_{b}x$.

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

$={\displaystyle \frac{{\log }_c(b^{{\log }_{b}x})}{{\log }_{c}b}}$

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

$\cos (\alpha +\beta )+i\sin (\alpha +\beta )=e^{i(\alpha +\beta )}$

$=e^{i\alpha }{e}^{i\beta }$

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

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

$\sin 2\theta =\sin (\theta +\theta )$

$=\sin \theta \cos \theta +\cos \theta \sin \theta$

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

$\cos 2\theta =\cos (\theta +\theta )$

$=\cos \theta \cos \theta -\sin \theta \sin \theta$

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

${\cos }^2\theta -{\sin }^2\theta =({\cos }^2\theta +{\sin }^2\theta )-2{\sin }^2\theta$

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

${\cos }^2\theta -{\sin }^2\theta =2{\cos }^2\theta -({\cos }^2\theta +{\sin }^2\theta )$

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

$\cos n\theta +i\sin n\theta =(\cos \theta +i\sin \theta {)}^n$

$={\displaystyle \sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k}){\cos }^{n-k}\theta i^k{\sin }^k\theta }$