# How can I understand and prove the "sum and difference formulas" in trigonometry?

The "sum and difference" formulas often come in handy, but it's not immediately obvious that they would be true.

\begin{align} \sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\ \cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \end{align}

So what I want to know is,

1. How can I prove that these formulas are correct?
2. More importantly, how can I understand these formulas intuitively?

Ideally, I'm looking for answers that make no reference to Calculus, or to Euler's formula, although such answers are still encouraged, for completeness.

• – glS
Commented Apr 29, 2019 at 11:16
• What does the upside down plus or minus symbol that appears somewhere in your question actually mean? Was it just a typo? Commented Jun 1, 2019 at 3:55
• @Timothy not a typo. See en.wikipedia.org/wiki/Plus-minus_sign Commented Jul 22, 2019 at 1:20

Here are my favorite diagrams:

As given, the diagrams put certain restrictions on the angles involved: neither angle, nor their sum, can be larger than 90 degrees; and neither angle, nor their difference, can be negative. The diagrams can be adjusted, however, to push beyond these limits. (See, for instance, this answer.)

Here's a bonus mnemonic cheer (which probably isn't as exciting to read as to hear):

Sine, Cosine, Sign, Cosine, Sine!
Cosine, Cosine, Co-Sign, Sine, Sine!

The first line encapsulates the sine formulas; the second, cosine. Just drop the angles in (in order $$\alpha$$, $$\beta$$, $$\alpha$$, $$\beta$$ in each line), and know that "Sign" means to use the same sign as in the compound argument ("+" for angle sum, "-" for angle difference), while "Co-Sign" means to use the opposite sign.

• BTW: Apparently, someone copped my image and used an adapted version (plus a tangent rule variant) in the Wikipedia "List of Trigonometric Identities" page. I'm flattered, of course, but I don't think I was given proper credit according to the Creative Commons Attribution Share Alike license, or StackExchange's Terms of Service. Actually, I would've appreciated if the Wikipedia contributor had simply invited me to submit the image myself. (A note about the whole thing would've been nice.)
– Blue
Commented Aug 20, 2013 at 18:05
• I have now sourced you on Wikipedia. I have also asked you to be sourced in future from the original uploader. Commented Nov 28, 2013 at 12:27
• FYI: These diagrams now appear on my Trigonography website.
– Blue
Commented Nov 15, 2015 at 1:03
• @Blue - I have been looking ALL OVER for a visual of cos(a - b). This is the ONLY one I've been able to find. Thank you so much for taking the time to create this diagram and posting it! This is extremely helpful! Commented Apr 30, 2020 at 12:09
• Shameless plug: I've put a modified version of the design onto a t-shirt, via Teespring. (I'll delete this comment if there are objections ... so, I guess there is some shame involved.)
– Blue
Commented Mar 21, 2021 at 11:11

The key fact here is that rotation is a linear transformation, e.g. the rotation of $u + v$ is the rotation of $u$ plus the rotation of $v$. You should draw a diagram that shows this carefully if you don't believe it. That means a rotation is determined by what it does to $(1, 0)$ and to $(0, 1)$.

But $(1, 0)$ rotated by $\theta$ degrees counterclockwise is just $(\cos \theta, \sin \theta)$, whereas $(0, 1)$ rotated by $\theta$ degrees counterclockwise is just $(-\sin \theta, \cos \theta)$. (Again, draw a diagram.) That means a rotation by $\theta$ is given by a $2 \times 2$ matrix with those entries. (Matrices don't work here yet.)

So take a rotation by $\theta$ and another one by $\theta'$, and multiply the corresponding matrices. What you get is the sine and cosine angle addition formulas. (The connection to complex numbers is that one can represent complex numbers as $2 \times 2$ real matrices.)

Also, if you believe that $a \cdot b = |a| |b| \cos \theta$, this implies the cosine angle difference formula when $a$ and $b$ are unit vectors. Ditto for the cross product and the sine angle difference formula.

• I might argue that this doesn't answer 1.... how to prove these formulas, unless you show generally just with brute euclidean geometry that rotation is linear...?
– SBK
Commented May 30, 2018 at 22:07
• not the most rigorous, but certainly intuitive, and the one i am going to use on the go from now on thanks Commented Jan 7, 2022 at 18:35
• I propose a graphical analogy/supplement to this answer in an answer of my own, although I think the matrix approach used here is still a neater approach. Commented Aug 31, 2022 at 2:22
• (+1) This question was referenced by another question, and I didn't see any other matrix-based answers, so I posted an answer. I now see that this is what you had proposed, but it seems MathJax was not up to the task when you posted.
– robjohn
Commented Jun 12, 2023 at 8:49

Though the standard high-school derivations are not the most useful way to remember it in the long run, here's another one which I like because you can "see" it directly without much algebra.

Let P be the point on the unit circle got by rotating (1,0) by angle α+β. Drop a perpendicular N to the α-rotated line, and R to the x-axis. So from the right triangle ONP, you see ON = cos β. You can see that the angle RPN is α too: it's the complement of ∠PNQ, and so is ∠QNO = α. Now,

$\sin(\alpha + \beta) = \mbox{PR} = \mbox{PQ} + \mbox{QR} = \sin(\beta)\cos(\alpha) + \cos(\beta)\sin(\alpha)$, and

$\cos(\alpha + \beta) = \mbox{OR} = \mbox{OM} - \mbox{RM} = \cos(\beta)\cos(\alpha) - \sin(\beta)\sin(\alpha)$.

• with program did you draw this diagram? Commented Jul 13, 2014 at 17:33
• @Assad: If I remember correctly, I used TikZ, and this was in fact my first time using TikZ. I wish I had kept the source code of this figure; I haven't used TikZ much since then, and I'd have to re-learn it if I wanted to draw this again from scratch. :-) But it couldn't have been too hard, because I did learn enough to draw this. Commented Jul 13, 2014 at 17:45
• This is the best answer because it's the simplest (not using imaginary numbers) and has the best visual. Commented May 16, 2018 at 3:12

You can use the complex representation,
$\cos(x) = \frac{1}{2}(e^{ix} + e^{-ix})$
$\sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$
and the rules for powers ($a^{x+y}=a^x a^y$)

• This is equivalent to diagonalizing the rotation matrices before multiplying them. Commented Jul 31, 2010 at 17:35
• @QiaochuYuan Indeed, but in fewer words ;) Commented Oct 6, 2014 at 6:47

I will prove the identity $\cos(x+y)=\cos x\cos y-\sin x\sin y$, using with the following definitions of sine and cosine:

$$\sin x:= \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!} \ \ \ \ ;\ \ \ \cos x:= \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$$

Proof:

$$\cos (x+y)= \sum_{n=0}^{\infty}(-1)^n\frac{(x+y)^{2n}}{(2n)!}$$

Using the Binomial theorem, we will have

$$\sum_{n=0}^{\infty}(-1)^n\sum^{2n}_{k=0}\binom{2n}{k}\frac{x^ky^{2n-k}}{(2n)!}=$$ $$=\sum_{n=0}^{\infty}(-1)^n\sum^{2n}_{k=0}\frac{x^ky^{2n-k}}{k!(2n-k)!}$$

Now, separating the inner sum into two, for even $k$ and for odd $k$:

$$=\sum_{n=0}^{\infty}(-1)^n\sum^{n}_{k=0}\frac{x^{2k}y^{2n-2k}}{(2k)!(2n-2k)!}+\sum_{n=1}^{\infty}(-1)^n\sum^{n-1}_{k=0}\frac{x^{2k+1}y^{2n-2k-1}}{(2k+1)!(2n-2k-1)!}$$

Now, let us look on the first sum,

$$\sum_{n=0}^{\infty}(-1)^n\sum^{n}_{k=0}\frac{x^{2k}y^{2n-2k}}{(2k)!(2n-2k)!}=$$ $$=\sum_{n=0}^{\infty}\sum^{n}_{k=0}(-1)^k\frac{x^{2k}}{(2k)!}(-1)^{n-k}\frac{y^{2(n-k)}}{(2(n-k))!}=$$

By Cauchy product, we have:

$$=\sum_{n=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}\sum_{n=0}^{\infty}(-1)^k\frac{y^{2k}}{(2k)!}=$$

$$=\cos x\cos y$$

For the second sum,

$$\sum_{n=1}^{\infty}(-1)^n\sum^{n-1}_{k=0}\frac{x^{2k+1}y^{2n-2k-1}}{(2k+1)!(2n-2k-1)!}=$$

By Cauchy product, we have:

$$\sum_{n=1}^{\infty}\sum^{n-1}_{k=0}(-1)^k\frac{x^{2k+1}}{(2k+1)!}(-1)^{n-k}\frac{y^{2((n-1)-k)+1}}{(2((n-1)-k)+1)!}$$

And by substituting $t=n-1$, we will have:

$$\sum_{t=0}^{\infty}\sum^{t}_{k=0}(-1)^k\frac{x^{2k+1}}{(2k+1)!}(-1)^{t+1-k}\frac{y^{2(t-k)+1}}{(2(t-k)+1)!}$$

$$=-[ \sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!} ][\sum_{k=0}^{\infty}(-1)^k\frac{y^{2k+1}}{(2k+1)!}]$$

$$=-\sin x\sin y$$

Q.E.D

There are several typical derivations used in high school texts. Here's one:

diagram http://www.imgftw.net/img/400545892.png

Take two points on the unit circle, one a rotation of (1,0) by α, the other a rotation of (1,0) by β. Their coordinates are as shown in the diagram. Let c be the length of the segment joining those two points. By the Law of Cosines (on the blue triangle), $$c^2=1^2+1^2-2\cdot1\cdot1\cdot\cos(\alpha-\beta)$$. Using the distance formula, $$c=\sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}$$. Squaring the latter and setting the two equal, $$1^2+1^2-2\cdot1\cdot1\cdot\cos(\alpha-\beta)=(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2$$. Simplifying both sides, $$2-2\cos(\alpha-\beta)=\cos^2\alpha-2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta$$ $$=2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$$ (using the Pythagorean identity). Solving for $$\cos(\alpha-\beta)$$, $$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$.

From this identity, the other three can be derived by substituting $$\frac{\pi}{2}-\alpha$$ for α (gives sin(α+β)), then -β for β (gives the remaining two).

As to understanding the formulas intuitively if you accept that multiplying by a complex number $$z_\theta$$ for which |z|=1 rotates by θ, then you can think about what happens when you multiply $$z_\alpha=\cos\alpha+i\sin\alpha$$ and $$z_\beta=\cos\beta+i\sin\beta$$ (by expanding the binomial product), which should result in $$\cos(\alpha+\beta)+i\sin(\alpha+\beta)$$.

• Point worth making: Isaac's last paragraph and my argument are the same. This is a point which is not often understood. Commented Jul 31, 2010 at 7:53
• @Qiaochu Yuan: Yes, quite true. I think about it in complex numbers more naturally than in matrices, but it's equivalent. I don't think of it as a proof because my chain of derivations usually uses the sum/difference identities to justify that complex multiplication (by a number of modulus 1) is geometrically a rotation. Commented Jul 31, 2010 at 7:55
• Ah. You don't need to do that: you just need to construct the isomorphism between 2x2 rotation matrices and the complex numbers. Commented Jul 31, 2010 at 8:02
• @Isaac: You don't even need to involve rotation matrices. There is a wonderfully simple picture which explains why multiplication by $w$ scales the plane by $|w|$ and rotates by $\arg w$. I can't draw it here, but it's Figure 6bc on p. 9 in Needham's Visual Complex Analysis. (You can view it on Google Books.) Commented Sep 3, 2010 at 6:58
• your diagram is gone Commented Oct 19, 2010 at 8:20

I remember that $e^{i\alpha}=\cos\alpha+i\sin\alpha$ and that $i^2=-1$. Both these relations are useful in many other situations and pretty fundamental to understanding complex numbers. Then your equalities are the real and, respectively, the imaginary part of $e^{i(\alpha+\beta)}=e^{i\alpha}e^{i\beta}$.

This is not very different from the other answers, but I actually prefer the algebra perspective. The only place where I think geometrically is in interpreting $e^{i\alpha}=\cos\alpha+i\sin\alpha$ by thinking of the unit circle in the complex plane.

• I prefer the geometric perspective too, and it just seems like you shouldn't need to know anything about imaginary numbers to understand these identities. Which is why I asked the question. The big insight I'm getting of course is that the two ways of looking at it are really not that different. Thanks! Commented Jul 31, 2010 at 23:14

Another method puts the sine double angle formula in a similar framework with the property $$e^{x+y} = e^x e^y$$ (surprise!) and generalizes to give identities for elliptic functions. (This derivation of the sine addition formula is valid just for restricted $$\alpha$$, and $$\beta$$, as are some of the geometric arguments given by others.)

The idea: first prove the identity $$\int_0^x \frac{dt}{\sqrt{1-t^2}} + \int_0^y \frac{dt}{\sqrt{1-t^2}} = \int_0^{T(x,y)} \frac{dt}{\sqrt{1-t^2}} \qquad \qquad (1)$$ for all pairs of real numbers $$x$$ and $$y$$ for which $$x^2 + y^2 < 1$$, where $$T(x,y) = x\sqrt{1-y^2} + y\sqrt{1-x^2}$$. This identity says $$\arcsin(x) + \arcsin(y) = \arcsin(T(x,y))$$ and then setting $$x=\sin(u)$$ and $$y=\sin(v)$$ gives $$\sin(u+v) = T(\sin(u),\sin(v)) = \sin(u) \cos(v) + \cos(u) \sin(v)$$ (for certain restricted $$u$$, $$v$$ -- for instance, when $$u$$, $$v \in \left( -\frac{\pi}{4}, \frac{\pi}{4} \right)$$). To prove Identity (1), consider $$y$$ as fixed and show that both sides have the same derivative. That is, show that $$\frac{1}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-(T(x,y))^2}} \cdot \frac{d}{dx} T(x,y)$$ This identity can be proved with algebra. Then both sides of (1) differ by a constant, and evaluating at $$x=0$$ shows that they are actually equal.

A simpler argument proves the identity $$\int_1^x \frac{dt}{t} + \int_1^y \frac{dt}{t} = \int_1^{xy} \frac{dt}{t},$$ from which we obtain the addition property for natural logarithms, and then by inverting, the identity $$e^{x+y} = e^x e^y$$.

Now Leonard Euler considered the function $$F(x) = \int_0^x \frac{dt}{\sqrt{1-t^4}}.$$ It can be shown that this function cannot be written in terms of the standard list of common functions. Using similar reasoning to above (but with much more complicated algebra after taking the derivative) one can show $$F(x) + F(y) = F(T(x,y))$$ where now $$T(x,y) = \frac{x\sqrt{1-y^4} + y\sqrt{1-x^4}}{x^2y^2}$$ Thus, we obtain another addition formula. Perhaps by analogy with the above, Abel inverted the function $$F$$, which allowed him to rewrite the property as an "addition formula" for the inverse of $$F$$ analogous to the sine addition formula. This idea leads to elliptic functions, which in turn led to the modern theory of elliptic curves.

• This is a late comment, but what is $T$? Commented Jun 10, 2021 at 10:03
• Thanks. I had defined the second, more complicated T at the bottom, but I forgot to define the one at the top! It's there now. Commented Jun 11, 2021 at 11:40

Consider a unit circle with $$O$$ as the centre. Let $$P_{1}$$, $$P_2$$ and $$P_{3}$$ be points on the circle making angles $$A$$, $$B$$ and $$A−B$$, respectively, with the positive direction of the X-axis. We know that if two chords subtend equal angle at the centre, then the chords are equal and chords $$P_{3}P_{0}$$ and $$P_1P_2$$ subtend equal angles at $$O$$. Therefore, $$\overline{P_{3}P_{0}}=\overline{P_1P_2}$$. By distance formula, the distance between the points $$P_{0}(1,0)$$ and $$P_{3}(\cos(A−B),\sin(A−B))$$ is $$P_{3}P_{0}=\sqrt{\left(\cos \left( A-B \right) - 1\right) ^{2} + \left(\sin \left( A-B \right) - 0\right) ^{2}}$$ Similarly, the distance between the points $$P_{1}(\cos A,\sin A)$$ and $$P_{2}(\cos B,\sin B)$$ is $$P_1P_2= \sqrt{\left( \cos B-\cos A\right) ^{2}+\left( \sin B - \sin A \right) ^{2}}$$

On squaring both sides, we get

$$\begin{array}{ll} \Rightarrow & \left\{\cos (A-B)-\left.1\right|^{2}+\sin ^{2}(A-B)=(\cos B-\cos A)^{2}+(\sin B-\sin A)^{2}\right. \\ \Rightarrow & \cos ^{2}(A-B)-2 \cos (A-B)+1+\sin ^{2}(A-B)=\cos ^{2} B+\cos ^{2} A-2 \cos A \cos B \\ & +\sin ^{2} B+\sin ^{2} A-2 \sin A \sin B \\ \Rightarrow & 2-2 \cos (A-B)=2-2 \cos A \cos B-2 \sin A \sin B \\ \Rightarrow & \cos (A-B)=\cos A \cos B+\sin A \sin B \\ \text { Hence, } & \cos (A-B)=\cos A \cos B+\sin A \sin B \end{array}$$

• Don't know why this doesn't have more upvotes. This is the only completely correct answer following the definitions in the question Commented May 10, 2023 at 15:48

The following is (essentially) a graphical version of an earlier answer, relying on some simple vector operations in two dimensions but without explicitly invoking matrices.

The argument is summarized in the following diagram.

This is meant to be a proof without words, but for clarity, I'll build up to the conclusion step by step below.

First, let's just consider the unit circle in the usual Cartesian plane. Here it is, along with the $$x$$ and $$y$$ axes and unit vectors from the origin to the two points $$(1,0)$$ and $$(0,1).$$ We use $$A=(1,0)$$ and $$B=(0,1)$$ to denote the position vectors of these two points.

Next we add a point at an angle $$\alpha$$ counterclockwise around the circle from the point $$A=(1,0).$$ According to the unit circle interpretation of the sine and cosine, the coordinates of this point, and hence the coordinates of its position vector, are $$C = (\cos\alpha, \sin\alpha).$$

But as illustrated in by the vector sum figure, this is equivalent to saying that $$C$$ is a linear combination of the two vectors $$A=(1,0)$$ and $$B=(0,1),$$

$$C = (\cos\alpha) A + (\sin\alpha) B.$$

The equation above is a coordinate-free way of constructing a unit position vector rotated an angle $$\theta$$ counterclockwise from a given unit vector: find a unit vector at a right angle counterclockwise from the given vector, and the rotated vector is $$\cos\theta$$ of the given vector combined with $$\sin\theta$$ of the right-angle vector.

Now that we have the unit vector $$C$$, we construct a unit vector $$D$$ at a right angle clockwise from $$C$$:

The vector sum figure, rotated $$90$$ degrees from the figure in the previous diagram, shows that the $$x$$ coordinate of $$D$$ is the opposite of the $$y$$ coordinates of $$C$$ while the $$y$$ coordinate of $$D$$ is simply the $$x$$ coordinates of $$C.$$ If this is not obvious enough, you can verify it for a vector $$C$$ in each of the four quadrants.

Now we construct a unit vector $$E$$ at an angle $$\beta$$ counterclockwise from $$C.$$

We construct this vector as a linear combination of the given vector $$C$$ and the vector $$D$$ at a right angle counterclockwise from $$C,$$ that is,

$$E = (\cos\beta) C + (\sin\beta) D.$$

Finally, we multiply the coordinates of $$C$$ by the scalar $$\cos\beta$$, multiply the coordinates of $$D$$ by the scalar $$\sin\beta$$, and add the resulting vectors coordinatewise to get the coordinates of $$E$$:

$$E = (\cos\beta\cos\alpha - \sin\beta\sin\alpha, \cos\beta\sin\alpha + \sin\beta\cos\alpha).$$

But $$E$$ is a vector at an angle $$\alpha + \beta$$ counterclockwise from $$A=(1,0),$$ and therefore its coordinates give two trigonometric functions of $$\alpha + \beta,$$

\begin{align} \cos(\alpha + \beta) &= \cos\beta\cos\alpha - \sin\beta\sin\alpha, \\ \sin(\alpha + \beta) &= \cos\beta\sin\alpha + \sin\beta\cos\alpha. \end{align}

Simple rearrangement within the terms gets us the usual formulas for these functions.

As can be seen in the diagram, since rotating $$\begin{bmatrix}1&0\end{bmatrix}$$ counterclockwise by $$\alpha$$ gives $$\begin{bmatrix}\cos(\alpha)&\sin(\alpha)\end{bmatrix}$$ and rotating $$\begin{bmatrix}0&1\end{bmatrix}$$ counterclockwise by $$\alpha$$ gives $$\begin{bmatrix}-\sin(\alpha)&\cos(\alpha)\end{bmatrix}$$, we get $$\begin{bmatrix}x&y\end{bmatrix}=\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\color{#C00}{\xrightarrow[\text{rotation by }\alpha]{\text{counterclockwise}} \begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}\cos(\alpha)&\sin(\alpha)\\-\sin(\alpha)&\cos(\alpha)\end{bmatrix}}\tag1$$ Note that counterclockwise rotation by $$-\alpha$$ is clockwise rotation by $$\alpha$$. That is, counterclockwise rotation by $$-\alpha$$ is the inverse of counterclockwise rotation by $$\alpha$$. To verify this, use $$\cos(-\alpha)=\cos(\alpha)$$, $$\sin(-\alpha)=-\sin(\alpha)$$, and matrix multiplication: \begin{align} &\overbrace{\begin{bmatrix} \cos(\alpha)&\sin(\alpha)\\ -\sin(\alpha)&\cos(\alpha) \end{bmatrix}}^{\substack{\text{counterclockwise}\\\text{rotation by }\alpha}} \overbrace{\begin{bmatrix} \cos(\alpha)&-\sin(\alpha)\\ \sin(\alpha)&\cos(\alpha) \end{bmatrix}}^{\substack{\text{counterclockwise}\\\text{rotation by }-\alpha}}\tag{2a}\\ &= \begin{bmatrix} \cos^2(\alpha)+\sin^2(\alpha)&0\\ 0&\cos^2(\alpha)+\sin^2(\alpha) \end{bmatrix}\tag{2b}\\ &= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}\tag{2c} \end{align}

Rotation by $$\alpha+\beta$$ is just rotation by $$\alpha$$ followed by rotation by $$\beta$$. Using matrix multiplication, we get \begin{align} &\begin{bmatrix} \cos(\alpha+\beta)&\sin(\alpha+\beta)\\ -\sin(\alpha+\beta)&\cos(\alpha+\beta) \end{bmatrix}\tag{3a}\\ &=\begin{bmatrix} \cos(\alpha)&\sin(\alpha)\\ -\sin(\alpha)&\cos(\alpha) \end{bmatrix} \begin{bmatrix} \cos(\beta)&\sin(\beta)\\ -\sin(\beta)&\cos(\beta) \end{bmatrix}\tag{3b}\\ &=\begin{bmatrix} \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)&\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\\ -\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)&\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \end{bmatrix}\tag{3c} \end{align} Equating elements between $$\text{(3a)}$$ and $$\text{(3c)}$$, we get \begin{align} \cos(\alpha+\beta)&=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\tag{4a}\\[6pt] \sin(\alpha+\beta)&=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\tag{4b} \end{align} Using $$\cos(-\alpha)=\cos(\alpha)$$ and $$\sin(-\alpha)=-\sin(\alpha)$$, we get \begin{align} \cos(\alpha-\beta)&=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)\tag{5a}\\[6pt] \sin(\alpha-\beta)&=\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)\tag{5b} \end{align}

You might take refuge to complex numbers and use the Euler relation $\exp(i\phi)=\cos(\phi)+i\sin(\phi)$ and the fundamental property of the $\exp$ function:

$\cos(\alpha+\beta)+i\sin(\alpha+\beta)=\exp(i(\alpha+\beta))=\exp(i\alpha)\cdot\exp(i\beta)=$
$=(\cos(\alpha)+i\sin(\alpha))\cdot(\cos(\beta)+i\sin(\beta))=$
$=(\cos(\alpha)\cdot\cos(\beta)-\sin(\alpha)\cdot\sin(\beta))+i(\cos(\alpha)\cdot\sin(\beta)+\sin(\alpha)\cdot\cos(\beta))$

Finally use therefrom the real resp. imaginary part separately.
This is how you'd get both the trigonometric addition theorems.

--- rk

I just recently came up with this, so I thought I'd share. By utilizing the complex plane, I can easily derive the double angle formulas in my head, and quickly develop the sum and difference formulas on paper. You need only understand that multiplying by a complex number amounts to scaling and rotation in the plane.

Let $$\alpha, \beta, \theta \in \mathbb{R}.$$ Consider vectors $$\vec{P} = (\cos(\alpha), \sin(\alpha)), \vec{Q} = (\cos(\beta), \sin(\beta) ),$$ and their rotated versions $$\vec{P'} = (\cos(\alpha + \theta), \sin(\alpha + \theta) ), \vec{Q'} = (\cos(\beta + \theta), \sin(\beta + \theta) ).$$

Rotations preserve distances, so $$PQ = P'Q',$$ ie $$(\cos(\alpha) - \cos(\beta)) ^2 + (\sin(\alpha) - \sin(\beta) ) ^2$$ $$= (\cos(\alpha + \theta) - \cos(\beta + \theta) ) ^2 + (\sin(\alpha + \theta) - \sin(\beta + \theta) ) ^2 ,$$ ie $${\color{green}{\cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta) = \cos(\alpha + \theta) \cos(\beta + \theta) + \sin(\alpha + \theta) \sin(\beta + \theta)}}.$$

This holds for all values of $$\alpha, \beta, \theta.$$
Setting $$\alpha = 0,$$ $$\cos(\beta) = \cos(\theta) \cos(\beta + \theta) + \sin(\theta) \sin(\theta + \beta).$$ Further setting $$\lbrace \beta = A+B, \theta = (-B) \rbrace$$ gives $$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B).$$

Similarly setting $$\alpha = \frac{\pi}{2}$$ gives $$\sin(\beta) = -\sin(\theta) \cos(\beta + \theta) + \cos(\theta) \sin(\beta + \theta),$$ and further setting $${ \lbrace \beta = A + B, \theta = (-B) \rbrace }$$ gives $${ \sin(A+B) = \sin(A) \cos(B) + \cos(A) \sin(B) }.$$

There are simple geometric proofs of the formulas for $$\sin(\alpha \pm \beta)$$ and $$\cos(\alpha \pm \beta)$$ for the case where $$\alpha,$$ $$\beta,$$ and $$\alpha \pm \beta$$ are all acute angles. The answer I am linking here is a great example.

To move beyond acute angles you need to have a notion of what the sines and cosines of non-acute angles are. One can extend the graphical proofs to other cases one by one, but wouldn't it be nice to cover all possible angles once and for all? So let's use a few additional facts that can be more or less read directly from the unit circle definition, for example $$\sin(\tfrac\pi2 - \theta) = \cos(\theta)$$ and $$\sin(\pi - \theta) = \sin(\theta).$$

If $$\alpha + \beta = \tfrac\pi2$$ then $$\beta = \tfrac\pi2 - \alpha,$$ \begin{align} \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) &= \sin(\alpha)\cos(\tfrac\pi2 - \alpha) + \cos(\alpha)\sin(\tfrac\pi2 - \alpha) \\ &= \sin^2(\alpha) + \cos^2(\alpha) \\ &= 1 = \sin(\alpha + \beta), \end{align} and \begin{align} \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) &= \cos(\alpha)\cos(\tfrac\pi2 - \alpha) - \sin(\alpha)\sin(\tfrac\pi2 - \alpha) \\ &= \cos(\alpha)\sin(\alpha) - \sin(\alpha)\cos(\alpha) \\ &= 0 = \sin(\alpha + \beta), \end{align} so the sum formulas both work in that case.

If $$\alpha + \beta$$ is an obtuse angle then \begin{align} \sin(\alpha)\cos&(\beta) + \cos(\alpha)\sin(\beta) \\ &= \cos(\tfrac\pi2 - \alpha)\sin(\tfrac\pi2 - \beta) + \sin(\tfrac\pi2 - \alpha)\cos(\tfrac\pi2 - \beta) \\ &= \sin((\tfrac\pi2 - \alpha) + (\tfrac\pi2 - \beta)) \\ &= \sin(\pi - (\alpha + \beta)) \\ &= \sin(\alpha + \beta) \\ \end{align} and \begin{align} \cos(\alpha)\cos&(\beta) - \sin(\alpha)\sin(\beta) \\ &= \sin(\tfrac\pi2 - \alpha)\sin(\tfrac\pi2 - \beta) - \cos(\tfrac\pi2 - \alpha)\cos(\tfrac\pi2 - \beta) \\ &= -\cos((\tfrac\pi2 - \alpha) + (\tfrac\pi2 - \beta)) \\ &= -\cos(\pi - (\alpha + \beta)) \\ &= \cos(\alpha + \beta), \\ \end{align} so both sum formulas work in that case too.

There are no other cases for the sum of two acute angles, but let's also consider the zero angle. If $$\alpha = 0$$ then \begin{align} \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) &= 0 \cdot \cos(\beta) + 1\cdot \sin(\beta) \\ &= \sin(\beta) \\ &= \sin(\alpha + \beta) \end{align} and \begin{align} \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) &= 1 \cdot \cos(\beta) + 0\cdot \sin(\beta) \\ &= \cos(\beta) \\ &= \cos(\alpha + \beta), \end{align} so the formulas work in that case, and by symmetry they also work when $$\beta = 0$$. Obviously they work when both angles are zero.

That covers all possibilities for the sum of angles $$\alpha$$ and $$\beta$$ where $$0 \leq \alpha \lt \tfrac\pi2$$ and $$0 \leq \beta \lt \tfrac\pi2.$$

To move beyond this, let me define an ad-hoc notation to help express some formulas more conveniently. The ad-hoc notation makes the formulas less cluttered and (I hope) easier to grasp. For arbitrary angles $$\theta,$$ define $$\sin_k(\theta) \triangleq \sin(\theta + k \cdot \tfrac\pi2).$$

Then \begin{align} \sin_0(\theta) &= \sin(\theta + 0 \cdot \tfrac\pi2) = \sin(\theta), \\ \sin_1(\theta) &= \sin(\theta + 1 \cdot \tfrac\pi2) = \cos(\theta), \\ \sin_2(\theta) &= \sin(\theta + 2 \cdot \tfrac\pi2) = -\sin(\theta), \\ \sin_3(\theta) &= \sin(\theta + 3 \cdot \tfrac\pi2) = -\cos(\theta) \\ \end{align} for any angle $$\theta$$ (not just acute angles), as one can easily verify using the unit circle, and the pattern repeats four lines at a time (either after or before this set of equations) by adding or subtracting $$2\pi$$ from each angle; $$\sin_{k+4}(\theta) = \sin_k(\theta) = \sin_{k-4}(\theta).$$ Also note that $$\sin_{k+2}(\theta) = -\sin_k(\theta) = \sin_{k-2}(\theta).$$

Then for any integers $$m$$ and $$n,$$ \begin{align} \sin_m(\alpha)\sin_{n+1}(\beta) &= (-\sin_{m+2}(\alpha))(-\sin_{n-1}(\beta)) \\ &= \sin_{m+2}(\alpha)\sin_{n-1}(\beta). \end{align}

It follows that \begin{align} \sin_m(\alpha)\sin_{n+1}&(\beta) + \sin_{m+1}(\alpha)\sin_n(\beta) \\ &= \sin_{m+1}(\alpha)\sin_n(\beta) + \sin_{m+2}(\alpha)\sin_{n-1}(\beta) \end{align} by increasing $$m$$ by $$2$$ and decreasing $$n+1$$ by $$2$$ in $$\sin_m(\alpha)\sin_{n+1}(\beta)$$ and then reversing the order of the sum. The result is that the "$$m$$" subscripts both increase by $$1$$ and the "$$n$$" subscripts both decrease by $$1.$$ We can repeat this set of operations as many times as we like, forward or backward, so for any integer $$k,$$ \begin{align} \sin_m(\alpha)\sin_{n+1}&(\beta) + \sin_{m+1}(\alpha)\sin_n(\beta) \\ &= \sin_{m+k}(\alpha)\sin_{n-k+1}(\beta) + \sin_{m+k+1}(\alpha)\sin_{n-k}(\beta). \end{align}

In particular, for $$k = n,$$ \begin{align} \sin_m(\alpha)\sin_{n+1}&(\beta) + \sin_{m+1}(\alpha)\sin_n(\beta) \\ &= \sin_{m+n}(\alpha)\sin_1(\beta) + \sin_{m+n+1}(\alpha)\sin_0(\beta). \end{align}

Now that we have all this machinery, let $$\alpha$$ and $$\beta$$ be any angles of any magnitudes, positive or negative. Write \begin{align} \alpha &= \bar\alpha + a\cdot \tfrac\pi2, \\ \beta &= \bar\beta + b\cdot \tfrac\pi2. \end{align}

Then $$\sin(\alpha + \beta) = \sin(\bar\alpha + \bar\beta + (a + b)\tfrac\pi2) = \sin_{a+b}(\bar\alpha + \bar\beta).$$ We can cover all possible values of $$\alpha$$ and $$\beta$$ in just four cases, depending on the equivalence class of $$a + b$$ modulo $$4.$$ In each case we can rely on the knowledge of the angle sum formulas for acute angles $$\bar\alpha$$ and $$\bar\beta.$$

Case $$a + b \equiv 0 \pmod 4$$:

\begin{align} \sin_{a+b}(\bar\alpha + \bar\beta) &= \sin(\bar\alpha + \bar\beta) = \sin(\bar\alpha)\cos(\bar\beta) + \cos(\bar\alpha)\sin(\bar\beta) \\ &= \sin_0(\bar\alpha)\sin_1(\bar\beta) + \sin_1(\bar\alpha)\sin_0(\bar\beta) \\ &= \sin_{a+b}(\bar\alpha)\sin_1(\bar\beta) + \sin_{a+b+1}(\bar\alpha)\sin_0(\bar\beta) \end{align}

Case $$a + b \equiv 1 \pmod 4$$:

\begin{align} \sin_{a+b}(\bar\alpha + \bar\beta) &= \cos(\bar\alpha + \bar\beta) = \cos(\bar\alpha)\cos(\bar\beta) - \sin(\bar\alpha)\sin(\bar\beta) \\ &= \sin_1(\bar\alpha)\sin_1(\bar\beta) + \sin_2(\bar\alpha)\sin_0(\bar\beta) \\ &= \sin_{a+b}(\bar\alpha)\sin_1(\bar\beta) + \sin_{a+b+1}(\bar\alpha)\sin_0(\bar\beta) \end{align}

Case $$a + b \equiv 2 \pmod 4$$:

\begin{align} \sin_{a+b}(\bar\alpha + \bar\beta) &= -\sin(\bar\alpha + \bar\beta) = -\sin(\bar\alpha)\cos(\bar\beta) - \cos(\bar\alpha)\sin(\bar\beta) \\ &= \sin_2(\bar\alpha)\sin_1(\bar\beta) + \sin_3(\bar\alpha)\sin_0(\bar\beta) \\ &= \sin_{a+b}(\bar\alpha)\sin_1(\bar\beta) + \sin_{a+b+1}(\bar\alpha)\sin_0(\bar\beta) \end{align}

Case $$a + b \equiv 3 \pmod 4$$:

\begin{align} \sin_{a+b}(\bar\alpha + \bar\beta) &= -\cos(\bar\alpha + \bar\beta) = -\cos(\bar\alpha)\cos(\bar\beta) + \sin(\bar\alpha)\sin(\bar\beta) \\ &= \sin_3(\bar\alpha)\sin_1(\bar\beta) + \sin_0(\bar\alpha)\sin_0(\bar\beta) \\ &= \sin_{a+b}(\bar\alpha)\sin_1(\bar\beta) + \sin_{a+b+1}(\bar\alpha)\sin_0(\bar\beta) \end{align}

Note that in all four cases we transformed $$\sin_{a+b}(\bar\alpha + \bar\beta)$$ to the same expression, so we can finish each case as follows: \begin{align} \sin_{a+b}(\bar\alpha)\sin_1&(\bar\beta) + \sin_{a+b+1}(\bar\alpha)\sin_0(\bar\beta) \\ &= \sin_a(\bar\alpha)\sin_{b+1}(\bar\beta) + \sin_{a+1}(\bar\alpha)\sin_b(\bar\beta) \\ &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \end{align}

Therefore $$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)$$ for all angles $$\alpha$$ and $$\beta.$$

That's one of the four angle-sum/difference formulas for sine and cosine. The others follow easily now that we know that the formula for $$\sin(\alpha + \beta)$$ is not limited to positive acute angles. For $$\sin(\alpha - \beta),$$ \begin{align} \sin(\alpha - \beta) &= \sin(\alpha + (-\beta)) \\ &= \sin(\alpha)\cos(-\beta) + \cos(\alpha)\sin(-\beta) \\ &= \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta). \end{align}

For the cosine formulas, \begin{align} \cos(\alpha \pm \beta) &= \sin(\alpha \pm \beta + \tfrac\pi2) = \sin((\alpha + \tfrac\pi2) \pm \beta) \\ &= \sin(\alpha + \tfrac\pi2)\cos(\pm\beta) + \cos(\alpha + \tfrac\pi2)\sin(\pm\beta) \\ &= \cos(\alpha)\cos(\pm\beta) - \sin(\alpha)\sin(\pm\beta) \\ &= \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta). \end{align}

That's not as many cases as one might have expected. I wonder if we can reduce it all down to one case by somehow observing that for acute angles $$\bar\alpha$$ and $$\bar\beta,$$ $$\sin_k(\bar\alpha + \bar\beta) = \sin_k(\bar\alpha)\sin_1(\bar\beta) + \sin_{k+1}(\bar\alpha)\sin_0(\bar\beta),$$ but I have not hit on a method of proving that except by the four cases.

I will provide another proof for $$\sin{(x+y)}$$ that is possible because, although traditionally presented in textbooks as a consequence of the angle sum identity $$\sin{(x+y)}$$, the double-angle formula for the sine, $$\sin{(2x)}=2\sin{(x)}\cos{(x)}$$, can be derived independently of it (see the wonderful proof without words of "Start wearing purple"). Then, as Blue has pointed out, the fact that supplementary angles have the same sine is an easy consequence of the double-angle formula.

The following proof is valid for $$0\le\alpha\le\pi$$, $$0\le\beta\le\dfrac{\pi}{2}$$, $$0\le\alpha+\beta\le\pi$$.

Theorem. The sum identity for sine states that

$$\sin{(\alpha+\beta)}=\sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}\sin{(\beta)}.\tag{1}$$

Proof. Suppose $$\triangle{ABC}$$ is a triangle with sides $$\overline{BC}=a$$, $$\overline{AC}=b$$ and $$\overline{AB}=c$$. Let $$\angle{BAC}=\alpha$$, $$\angle{CBA}=\beta$$ and $$\angle{ACB}=\gamma$$. Recalling that the area of a triangle can be expressed as $$\Delta=\frac12bc\sin{(\alpha)}$$ and substituting from the cosine rule on the right-hand side of $$(1)$$, we have
\begin{aligned} \sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}\sin{(\beta)} &= \left(\frac{2\Delta}{bc}\right)\cos{(\beta)}+\left(\frac{2\Delta}{ac}\right)\cos{(\alpha)}\\&=\frac{2\Delta}{c}\left(\frac{\cos{(\beta)}}{b}+\frac{\cos{(\alpha)}}{a}\right)\\ &=\frac{2\Delta}{c}\left(\frac{a^2+c^2-b^2}{2abc}+\frac{b^2+c^2-a^2}{2abc}\right)\\&=\frac{2\Delta}{ab}\\&=\sin{(\gamma)}\\&=\sin{(\pi-(\alpha+\beta))}\\&=\sin{2\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}\\&=2\color{red}{\sin{\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}}\color{blue}{\cos{\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}}\\&=2\color{red}{\cos{\left(\frac{\alpha+\beta}{2}\right)}}\color{blue}{\sin{\left(\frac{\alpha+\beta}{2}\right)}}\\&= \sin{\left(\alpha+\beta\right)}.\end{aligned}

From a good old book, written by Professors at USNA ( Plane and spherical trigonometry by Lyman Kells et al.)