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.

10 Answers 10

up vote 47 down vote accepted

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 at 22:07

Here are my favorite diagrams:

Proof Without Words: Angle Sum and Difference for Sine and Cosine

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.

  • 1
    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
  • 11
    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
  • 2
    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 ($a^{x+y}=a^x a^y$)

  • 11
    This is equivalent to diagonalizing the rotation matrices before multiplying them. – Qiaochu Yuan Jul 31 '10 at 17:35
  • 2
    @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.

Angle sum formulae

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? – seeker Jul 13 '14 at 17:33
  • 1
    @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
  • This is the best answer because it's the simplest (not using imaginary numbers) and has the best visual. – user766353 May 16 at 3:12

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

  • 2
    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
  • 1
    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
  • 1
    @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
  • 5
    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.

  • You mean times. – Qiaochu Yuan Jul 31 '10 at 17:35
  • Sorry :( Fixed. Thanks. – rgrig Jul 31 '10 at 20:39
  • 1
    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

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. Click here for a pdf of my derivation

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 $x^2 + y^2 < 1$. 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.

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

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.