Trigonometric identities

Same angles:
$\alpha=\beta+k*360$

Opposite angles:
$\alpha$ and $-\alpha+k*360$

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

$\tan(-\alpha)=-\tan(\alpha)$
$\cot(-\alpha)=-\cot(\alpha)$

Supplementary angles:
$\alpha$ and $(180^\circ-\alpha)+k*360$

$\cos(180^\circ-\alpha)=-\cos(\alpha)$
$\sin(180^\circ-\alpha)=\sin(\alpha)$

$\tan(180^\circ-\alpha)=-\tan(\alpha)$
$\cot(180^\circ-\alpha)=-\cot(\alpha)$

Complementary angles:
$\alpha$ and $(90^\circ -\alpha)+k*360$

$\cos(90^\circ-\alpha)=\sin(\alpha)$
$\sin(90^\circ-\alpha)=\cos(\alpha)$

$\tan(90^\circ-\alpha)=\cot(\alpha)$
$\cot(90^\circ-\alpha)=\tan(\alpha)$

Anti-supplementary angles:
$\alpha$ and $(180^\circ +\alpha)+k*360$

$\cos(180^\circ+\alpha)=-\cos(\alpha)$
$\sin(180^\circ+\alpha)=-\sin(\alpha)$

$\tan(180^\circ+\alpha)=\tan(\alpha)$
$\cot(180^\circ+\alpha)=\cot(\alpha)$

Table

f(x)	0°	30°	45°	60°	90°
sin(A)	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1
cos(A)	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
tan(A)	0	$\frac{\sqrt{3}}{3}$	1	$\sqrt{3}$	und
cot(A)	und	$\sqrt{3}$	1	$\frac{\sqrt{3}}{3}$	0

$1+\tan^2(\alpha)=\sec^2(\alpha)$
$1+\cot^2(\alpha)=\csc^2(\alpha)$
Proof:
$1+\tan^2(\alpha)=1+\frac{\sin^2(\alpha)}{\cos^2(\alpha)}= \frac{\cos^2(\alpha)}{\cos^2(\alpha)}+\frac{\sin^2(\alpha)}{\cos^2(\alpha)}=\frac{1}{\cos^2(\alpha)}=\sec^2(\alpha)$
$1+\cot^2(\alpha)=1+\frac{\cos^2(\alpha)}{\sin^2(\alpha)}= \frac{\sin^2(\alpha)}{\sin^2(\alpha)}+\frac{\cos^2(\alpha)}{\sin^2(\alpha)}=\frac{1}{\sin^2(\alpha)}=\csc^2(\alpha)$

Angle sum and difference identities
$\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$
$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$
$\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$
$\sin(\alpha-\beta)=\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)$
$\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$
$\tan(\alpha-\beta)=\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$
Proof:

Remember that L²=(X2-X1)²+(Y2-Y1)²
The proof relies on the fact that W can be calculated 2 ways:
First way:
$P_1=(\cos\beta, \sin\beta)$
$P_2=(\cos\alpha, \sin\alpha)$
$W^2=(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2$

Second way:
$P_1=(\cos(\alpha-\beta), \sin(\alpha-\beta))$
$P_2=(1, 0)$
$W^2=(1-\cos(\alpha-\beta))^2+(0-\sin(\alpha-\beta))^2$

put them together:
$W^2=(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2=(1-\cos(\alpha-\beta))^2+(0-\sin(\alpha-\beta))^2$
Left side:
$W^2=\cos^2\alpha+\cos\beta^2-2\cos\alpha\cos\beta+\sin^2\alpha+\sin^2\beta-2\sin\alpha\sin\beta$
$W^2=\cos^2\alpha+\sin^2\alpha+\cos\beta^2+\sin^2\beta-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$
$W^2=1+1-2\cos\alpha\cos\beta-\sin\alpha\sin\beta$
$W^2=2-2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)$
Right side:
$W^2=1+\cos^2(\alpha-\beta)-2\cos(\alpha-\beta)+\sin^2(\alpha-\beta)$
$W^2=1+\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)-2\cos(\alpha-\beta)$
$W^2=1+1-2\cos(\alpha-\beta)$
$W^2=2-2\cos(\alpha-\beta)$
We can now clearly see that:
$2-2\cos(\alpha-\beta)=2-2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)$
or
$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$