# How can I understand and prove the “sum and difference formulas” in trigonometry? (cos(a ± b) = …, etc.)?

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.

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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.

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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.

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.

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Diagrams are neat; I had never seen them before. It was refreshing to see this related as relationships between triangles instead of relationships between unit circle angles. – Justin L. Aug 25 '10 at 3:09
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 Aug 20 '13 at 18:05
I have now sourced you on Wikipedia. I have also asked you to be sourced in future from the original uploader. – Chris Sherlock Nov 28 '13 at 12:27
FYI: These diagrams now appear on my Trigonography website. – Blue Nov 15 '15 at 1:03

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 ($a^{x+y}=a^x a^y$)

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This is equivalent to diagonalizing the rotation matrices before multiplying them. – Qiaochu Yuan Jul 31 '10 at 17:35
@QiaochuYuan Indeed, but in fewer words ;) – Tobias Kienzler Oct 6 '14 at 6:47

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

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

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with program did you draw this diagram? – seeker Jul 13 '14 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. – ShreevatsaR Jul 13 '14 at 17:45
Thank you, this looks like a neat application – seeker Jul 13 '14 at 17:47

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

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 $\cos(\alpha+\beta)+i\sin(\alpha+\beta)$.

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Point worth making: Isaac's last paragraph and my argument are the same. This is a point which is not often understood. – Qiaochu Yuan Jul 31 '10 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. – Isaac Jul 31 '10 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. – Qiaochu Yuan Jul 31 '10 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.) – Hans Lundmark Sep 3 '10 at 6:58
your diagram is gone – Tobias Kienzler Oct 19 '10 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.

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You mean times. – Qiaochu Yuan Jul 31 '10 at 17:35
Sorry :( Fixed. Thanks. – rgrig Jul 31 '10 at 20:39
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! – MatrixFrog Jul 31 '10 at 23:14

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

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