Proving sine of sum identity for all angles Could anyone present a proof of sine of sum identity for any pair of angles $a$, $b$?
$$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$
Most proofs are based on geometric approach (angles are $<90$ in this case). But please note the formula is supposed to work for any pair of angles.
The other derivation I know is using Euler's formula, namely this one.
There's one thing I don't feel comfortable with - we know that we add angles when multiplying two complex numbers. This is proven with sine of sum identity. So first we prove how multiplication of two complex exponentials works using sine of sum identity, and then use multiplication of complex exponentials to prove sine of sum identity. Can you tell me how it's not a circular argument?
 A: here is a geometric proof i saw in an old american mathematics monthly which uses the unit circle. first show that the square of the chord connecting $(1,0)$ and $(\cos t, \sin t)$ is $2(1-\cos t)$ using the distance formula. now reinterpret 
$$\text{ length of chord making an angle $t$ at the center is } 2 - 2\cos t $$
now compute the length squared between $\cos t, \sin t), (\cos s, \sin s)$ in two different ways:
(i) distance formula gives you $2 - \cos t \cos s - \sin t \sin s$ 
(ii) chord making an angle $t - s$ is $2 - \cos(t-s)$
equating the two gives you $$\cos (t-s) = \cos t \cos s + \sin t \sin s \tag 1$$ 
now use the fact $\cos \pi/2$ to derive  $\cos (\pi/2 - s) = \sin s$ by putting $t = \pi/2$ in $(1)$
put $t=0,$ to derive $\cos$ is an even function. put $t = -\pi/2,$ to show $\sin$ is an odd function. after all these you derive 
$$\sin(t-s) = \sin t \cos t - \cos t \sin s $$ and two for the sums.
A: You can prove $$e^{ix}=\cos(x)+i\sin(x)$$without using trig sum identities. e.g. let:$$y=\cos(x)+i\sin(x)\tag{1}$$$$\therefore \frac{dy}{dx}=-\sin(x)+i\cos(x)=iy$$$$\therefore \int\frac{1}{y}dy=\int idx$$$$\therefore \ln(y)=ix+C$$and we can show $C=0$ because from (1) $y=1$ when $x=0$, therefore:$$\ln(y)=ix$$$$\therefore y=e^{ix}$$that therefore removes the circular argument you mentioned in yur question.
A: Euler’s Formula...
$$ \sin(x) = \mbox{Im}(e^{ix}) = \frac{e^{ix} - e^{-ix}}{2i} $$
$$ \sin(a+b) = \mbox{Im}(e^{(a+b)i}) = \mbox{Im}(e^{ai} * e^{bi}) $$
$$ = \frac{e^{(a+b)i} - e^{-(a+b)i}}{2i} = \frac{(e^{ai} * e^{bi}) - (e^{-ai} * e^{-bi})}{2i} $$
$$ = \frac{((\cos(a)+i\sin(a))(\cos(b)+i\sin(b)) - ((\cos(-a)+i\sin(-a))(\cos(-b)+i\sin(-b))}{2i} $$
After some thorough simplifying...
$$ = \frac{(2i\sin(a)\cos(b) + 2i\cos(a)\sin(b))}{2i} $$
$$ \sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) $$
A: The statement:
$ e^{ix} = \sin(x) +i\cos(x) $
is derived using Taylor series which can be proven like so 
Just sum $\cos(x)$ and $\sin(x)$ for $f(x)$ to convince yourself. If you sum up those two Taylor series, it will give you the Taylor series for $e^{ix}$. 
We never aknowledged the existence of sum angle identity in this proof, hence there is not circular reasoning.
