# 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
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? Jun 1, 2019 at 3:55
• @Timothy not a typo. See en.wikipedia.org/wiki/Plus-minus_sign 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.

• 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. Aug 25, 2010 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, 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. Nov 28, 2013 at 12:27
• FYI: These diagrams now appear on my Trigonography website.
– Blue
Nov 15, 2015 at 1:03
• @Blue love your work! So much so, I offer constructive suggestions :-) I find it easy to get the "difference" version with $\cos-\beta=\cos\beta$ and $\sin-\beta=-\sin\beta$ - terms with $\sin\beta$ invert for the "difference" version, saving a diagram! I'm positive you don't want to do the diagrams again, but: I find it difficult to read vertical text. Khan Academy's version colours text depending on angle e.g. the whole of "$\cos\beta$" is red (matching the angle colour). Benefit: putting $\cos\beta$ first, you can see the triangle as having a unit hypotenuse, just scaled by $\cos\beta$. May 3, 2019 at 7:39

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...?
– T_M
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 Jan 7 at 18:35

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? 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. 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. 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. Jul 31, 2010 at 17:35
• @QiaochuYuan Indeed, but in fewer words ;) Oct 6, 2014 at 6:47

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. 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. 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. 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.) Sep 3, 2010 at 6:58
• your diagram is gone 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! Jul 31, 2010 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

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 $T$? 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. 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

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

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 $${ \sin(A+B) = \sin(A) \cos(B) + \cos(A) \sin(B) }.$$