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?
Answer
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.
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