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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?

Community
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user4205580
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    What is your definition of sine and cosine? – David H Jan 17 '15 at 23:52
  • The answer may depend on your definition of $\sin$, $\cos$. If you use the definition of $e^{ia}$ to be $\cos a + i \sin a$, then yes, the argument is circular. What is $e^z$? What is $\cos x$, $\sin x$? This needs to be sorted out. – orangeskid Jan 17 '15 at 23:53
  • I would use crossproduct but I guess it's the geometric approach (else this can be of course proven with analytical tools : derivatives and stuffs). – servabat Jan 17 '15 at 23:59
  • How many definitions of sine and cosine are there? Let $P=(x,y)$ be a point on cartesian plane, $r$ is the distance from the origin in the straight line. Then sine of the angle between $|OP|$ and positive $X$ is $\frac{y}{r}$ and cosine - $\frac{x}{r}$. – user4205580 Jan 18 '15 at 08:31
  • How many definitions? For any really interesting mathematical system, it is very likely that different people have constructed the definitions of that system in different (but mutually consistent) ways. Usually the $\theta$ in $\sin\theta$ and $\cos\theta$ is a number obtained in some way (such as the length of an arc of a unit circle). If it is defined as a particular geometric figure that must include the positive $x$-axis, then you have to define what you mean by "any angle" and "adding two angles". – David K Jan 18 '15 at 14:06

4 Answers4

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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.

abel
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  • I get the same result but shouldn't (i) be 2 - 2(cos t cos s + sin t sin s) and (ii) be 2 - 2cos(t-s)? – Cary Swoveland Apr 09 '25 at 20:46
  • I think the identity is written wrong in the last line; Sum difference identity is $sin(t-s)=sin(t)cos(s)-cos(t)sin(s)$. I am studying this in a pre calculus book, I may be wrong. I also would like to see more about how to go from the cos difference identity to the sin difference identity. I am lost after $sin(s)=cos(\frac{\pi}{2}-s), t = \frac{\pi}{2}$. If I rotate 1/2 around the unit circle something interesting happens. $sin(\frac{\pi}{2}=cos(0)$. But, sin is negative when I rotate to $sin(\frac{3\pi}{2}$. I am having a hard time accounting for the sign changes and +/- functions. – Trig Jun 03 '25 at 02:10
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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.

Mufasa
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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...

$$ = \frac{(2i\sin(a)\cos(b) + 2i\cos(a)\sin(b))}{2i} $$ $$ \sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) $$

Michael Lee
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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.

Daniele Tampieri
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GLaDOS
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