Using Euler's formula, it is possible to write the basic trig functions in terms of complex exponentials.  Cosine and sine are just the real and imaginary parts of the complex exponential

$$\displaystyle \Re \mbox{e} \left\{ e^{i\theta}\right\}=\frac{e^{i\theta}+e^{-i\theta}}{2}=\cos\theta\qquad\qquad\Im\mbox{m} \left\{ e^{i\theta}\right\}=\frac{e^{i\theta}-e^{-i\theta}}{2i}=\sin\theta$$

The tangent is then the ratio

$$\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{e^{i\theta}-e^{-i\theta}}{i\left(e^{i\theta}+e^{-i\theta}\right)}$$

The secant, cosecant, and cotangent are just the reciprocals

$$\displaystyle \sec\theta=\frac{2}{e^{i\theta}+e^{-i\theta}}\qquad\qquad\csc\theta=\frac{2i}{e^{i\theta}-e^{-i\theta}}\qquad\qquad\cot\theta=\frac{i\left(e^{i\theta}+e^{-i\theta}\right)}{e^{i\theta}-e^{-i\theta}}$$

The angle addition formulas can be derived easily using the complex exponential forms.  First let's look at angle addition formulas for sine and cosine

$$\displaystyle \sin(\alpha+\beta)=\Im\mbox{m}\left\{ e^{i(\alpha+\beta)}\right\} \qquad\qquad \cos(\alpha+\beta)=\Re\mbox{e} \left\{ e^{i(\alpha+\beta)}\right\}$$

noting that

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

$$\displaystyle {}=\cos\alpha\cos\beta-\sin\alpha\sin\beta+i\left(\cos\alpha\sin\beta+\sin\alpha\cos\beta\right)$$

the real and imaginary parts give the results

$$\displaystyle \cos\left(\alpha+\beta\right)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$

$$\displaystyle \sin\left(\alpha+\beta\right)=\cos\alpha\sin\beta+\sin\alpha\cos\beta$$

The angle difference (subtraction) formulas can be found by replacing $\beta$ with $-\beta$ in the above expressions and noting that cosine has even parity while sine has odd parity,

$$\displaystyle \cos\left(\alpha-\beta\right)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$

$$\displaystyle \sin\left(\alpha-\beta\right)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$$

The addition and subtraction formulas for tangent are simply the ratios of these

$$\displaystyle \tan\left(\alpha+\beta\right)=\frac{\cos\alpha\sin\beta+\sin\alpha\cos\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}=\frac{\sin\beta+\tan\alpha}{1-\tan\alpha\tan\beta}$$

$$\displaystyle \tan\left(\alpha-\beta\right)=\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$$

The double angle formulas can be found by setting $\beta=\alpha$ in the addition formulas

$$\displaystyle \cos2\alpha=\cos\left(\alpha+\alpha\right)=\cos^{2}\alpha-\sin^{2}\alpha$$

$$\displaystyle \sin2\alpha=\sin\left(\alpha+\alpha\right)=2\cos\alpha\sin\alpha$$

$$\displaystyle \tan2\alpha=\tan\left(\alpha+\alpha\right)=\frac{2\tan\alpha}{1-\tan^{2}\alpha}$$

The half-angle formulas can be found by using the Pythagorean identity to re-write the double angle formula for cosine in two different ways

$$\displaystyle \cos2\alpha=\cos^{2}\alpha-\sin^{2}\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$$

Then substituting $\beta=2\alpha$

$$\displaystyle \cos\beta=1-2\sin^2(\beta/2)$$

$$\displaystyle \cos\beta=2\cos^2(\beta/2)-1$$

The the half-angle formulas are then

$$\displaystyle \cos^{2}(\beta/2)=\frac{1+\cos\beta}{2}$$

$$\displaystyle \sin^{2}(\beta/2)=\frac{1-\cos\beta}{2}$$

$$\displaystyle \tan^{2}\left(\beta/2\right)=\frac{1-\cos\beta}{1+\cos\beta}$$

In general, multiple angle formulas can be computed from the real and imaginary parts of the complex exponential

$$\displaystyle e^{in\theta}=\left(e^{i\theta}\right)^{n}=\left(\cos\theta+i\sin\theta\right)^{n}$$

thus

$$\displaystyle \cos n\theta=\Re\mbox{e}\left\{ \left(\cos\theta+i\sin\theta\right)^{n}\right\}$$

and

$$\displaystyle \sin n\theta=\Im\mbox{m}\left\{ \left(\cos\theta+i\sin\theta\right)^{n}\right\}$$

For example, the half angle formula for cosine can be computed as

$$\displaystyle \cos\left(\alpha/2\right)=\Re\mbox{e}\left\{ \left(\cos\alpha+i\sin\alpha\right)^{1/2}\right\}=\frac{\left(\cos\alpha+i\sin\alpha\right)^{1/2}+\left(\cos\alpha-i\sin\alpha\right)^{1/2}}{2}$$

squaring both sides of the equation yields

$$\displaystyle \cos^{2}\left(\alpha/2\right)=\frac{\left[\left(\cos\alpha+i\sin\alpha\right)^{1/2}+\left(\cos\alpha-i\sin\alpha\right)^{1/2}\right]^{2}}{4}$$

which, upon expanding the square on the right hand side and simplifying, becomes

$$\displaystyle \cos^{2}\left(\alpha/2\right)=\frac{2+2\cos\alpha}{4}=\frac{1+\cos\alpha}{2}$$

which is the same as the result above.

Using the complex form of the trigonometric functions also allows us to write explicit representations for the inverse trigonometric functions.