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) 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:

Proof Sum & Difference Identity

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