# Trigonometry Identity: Prove that $\sin(a-b)=\sin a \cos b - \cos a \sin b$

First, I do not want a proof using $\sin(a+b)=\sin a \cos b + \cos a \sin b$. Second, I suspect that it has something to do with Euler's formula; $e^{ix}=\cos x + i\sin x$, but I am not sure. Can anyone give me some direction? Thanks in advance.

P.S. I apologize if this question is a duplicate. I did not find any when scrolling down the "Questions that may already have your answer" list.

• What's your definition of sine? – David H Mar 9 '14 at 7:30
• @DavidH I think we can assume that he means the trigonometric function. – William Chang Mar 9 '14 at 20:23
• @DavidH $\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}\dots$ (Taylor series), OR $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$. Does that answer your question – TrueDefault Mar 9 '14 at 20:26
• @Mathster You don't say? :p My point is that there half a dozen different routes for defining the trig functions, and the appropriate way to prove the identity in question depends on the definition you go with. That's why there's such wide variety of answers posted below. – David H Mar 9 '14 at 20:27
• @DavidH Ohh, okay haha. I was wondering what you meant. – William Chang Mar 9 '14 at 20:31

• Not that is cool! +1! – Robert Lewis Mar 9 '14 at 7:48
• okay you totally win this round. I was proud of the multiplying $e^{ix}$ and $e^{iy}$. This steals the show. – Guy Mar 9 '14 at 7:53
• It's a good graphic for the angle addition formula for $\sin$, but OP asks to not use the angle addition formula while proving the angle subtraction formula for $\sin$. Can you make an image like this with $\sin(\alpha-\beta)$? – alex.jordan Mar 9 '14 at 8:22
• This also only shows for $\alpha + \beta \in [0,\pi /2]$. – R R Mar 9 '14 at 8:28
• @alex.jordan: Done. – Lucian Mar 9 '14 at 8:51

The simplest way of deriving/proving this is through Euler's formula:

Take $e^{i(\theta+\alpha)}$ which, through basic exponent rules we know is equal to $e^{i\theta +i\alpha} = e^{i\theta}\cdot e^{i\alpha}\ .$ Now expanding this (using Euler's formula) we get: \begin{align} e^{i\theta}\cdot e^{i\alpha} &= (\cos(\theta)+i\sin(\theta))\cdot(\cos(\alpha)+i\sin(\alpha))\\ &= \cos(\theta)\cos(\alpha)-\sin(\theta)\sin(\alpha) + i(\cos(\theta)\sin(\alpha)+\sin(\theta)\cos(\alpha))\ \\ \end{align} and \begin{align} e^{i(\theta + \alpha)} = \cos(\theta + \alpha) + i \sin(\theta + \alpha) \end{align}\ . Remembering that $e^{i(\theta +\alpha)} = e^{i\theta}\cdot e^{i\alpha}\$we can see that $$\cos(\theta)\cos(\alpha)-\sin(\theta)\sin(\alpha) + i(\cos(\theta)\sin(\alpha)+\sin(\theta)\cos(\alpha))\ = \cos(\theta + \alpha) + i \sin(\theta + \alpha)$$

and because we know that the real part has to equal the real part and the complex has to equal the complex we get two equations: \begin{align} \cos(\theta)\cos(\alpha)-\sin(\theta)\sin(\alpha) &= \cos(\theta + \alpha)\\ \cos(\theta)\sin(\alpha)+\sin(\theta)\cos(\alpha) &= \sin(\theta + \alpha)\ . \end{align}

Now we simply make $\alpha = -\phi$ and remember that $\sin(-\theta) = -\sin(\theta)$ and $\cos(-\theta) = \cos(\theta)$, to get our subtraction identities: \begin{align} \cos(\theta - \phi) &=\cos(\theta)\cos(-\phi)-\sin(\theta)\sin(-\phi) \\ &= \cos(\theta)\cos(\phi)+\sin(\theta)\sin(\phi)\\ \\ \sin(\theta -\phi ) &= \cos(\theta)\sin(-\phi)+\sin(\theta)\cos(-\phi) \\ &= -\cos(\theta)\sin(\phi)+\sin(\theta)\cos(\phi)\ \ \blacksquare \end{align}

Now I will provide my favorite proof of this identity, which i consider more intuitive than the one above.

First we construct three right triangles, with two of them placed so that the hypotenuse of the first one is congruent and adjacent to the base of the other, and the third is constructed from the top point of the second to the base of the first (perpendicular to it):

It is clear from this construction that we are looking for the $\sin(\beta + \delta)$, which is equal to $\frac{\overline{DE}}{\overline{DA}}$. We however want to put this into terms of our first two triangles, which we presumably know more about. Because the identity we are looking for is actually a circular identity, or rather is true for the unit circle, we will let $\overline{DA}$ be a radius of the unit circle (i.e with a measure of $1$). This further reduces $\frac{\overline{DE}}{\overline{DA}}$ to $\frac{\overline{DE}}{1}$, meaning $$\sin(\beta + \delta) = \overline{DE}\ .$$

We can now construct a line from point $C$ that is perpendicular to $\overline{DE}$, in order to divide up line $\overline{DE}$:

Now we can say that because $\overline{DE} = \overline{DF} + \overline{FE}$, and $\overline{FE} \cong \overline{CB}$ that $$\sin(\beta + \delta) = \overline{DE} = \overline{DF} + \overline{CB}\ \ .$$

Now we try to find $\overline{CB}$ in terms of triangles we know. Because we know that $\sin(\delta) = \frac{\overline{CB}}{\overline{CA}}$ we can say that $\overline{CB} =\sin(\delta)\cdot \overline{CA}$. Now we can try to find $\overline{CA}$; noticing that $\cos(\beta) = \frac{\overline{CA}}{DA}$, but because $\overline{DA} = 1$ we say $\cos(\beta) = \overline{CA}$, we can write $\overline{CB} =\sin(\delta)\cos(\beta)$. Finally substituting into our equation for $\sin(\delta + \beta)$ we get $$\sin(\delta + \beta) = \overline{DF} + \sin(\delta)\cos(\beta)\ .$$

Finally we must find $\overline{DF}$, which can be written in terms of the angle $\angle FDC$ as $\cos(\angle FDC) \cdot \overline{DC} = \overline{DF}$. We know that $\angle FDC = 90^{\bigcirc} - \angle DCF$, and we know that $\angle DCF = 90^{\bigcirc} - \angle FCA$, as well as that $\angle FCA = 90^{\bigcirc} - \angle ACB$, and finally that $\angle ACB = 90^{\bigcirc} - \delta$. Now substituting appropriatly we get $$\angle FDC = \delta\ .$$ Which means that we can express $\cos(\angle FDC) \cdot \overline{DC} = \overline{DF}$ in terms of delta as follows $\overline{DF} = \cos(\delta) \cdot \overline{DC}\$ Because we know that $\overline{DC} = \sin(\beta)$ we can say $$\overline{DF} = \cos(\delta) \cdot \sin(\beta)\ .$$

Substituting into our $\sin(\delta + \beta)$ equation we get $$\sin(\delta + \beta) = \cos(\delta)\sin(\beta)+ \sin(\delta)\cos(\beta)\ .$$

Now we do as we did in our last proof and let $\beta = - \phi$ to get \begin{align} \sin(\delta -\phi ) &= \cos(\delta)\sin(-\phi)+\sin(\delta)\cos(-\phi) \\ &= -\cos(\delta)\sin(\phi)+\sin(\delta)\cos(\phi)\ \ \blacksquare \end{align}

EDIT: I just noticed you do not want arguments that use $\sin(\theta + \alpha)$, therefore I am going to make the first argument without $\sin(\theta + \alpha)$.

Take $e^{i(\theta-\alpha)}$ which, through basic exponent rules we know is equal to $e^{i\theta - i\alpha} = e^{i\theta}\cdot e^{-i\alpha}\ .$ Now expanding this (using Euler's formula) we get: \begin{align} e^{i\theta}\cdot e^{-i\alpha} &= (\cos(\theta)+i\sin(\theta))\cdot(\cos(-\alpha)+i\sin(-\alpha))\\ &= \cos(\theta)\cos(-\alpha)-\sin(\theta)\sin(-\alpha) + i(\cos(\theta)\sin(-\alpha)+\sin(\theta)\cos(-\alpha))\ \\ \end{align} and \begin{align} e^{i(\theta - \alpha)} = \cos(\theta - \alpha) + i \sin(\theta - \alpha) \end{align}\ . Remembering that $e^{i(\theta - \alpha)} = e^{i\theta}\cdot e^{-i\alpha}\$we can see that $$\cos(\theta)\cos(-\alpha)-\sin(\theta)\sin(-\alpha) + i(\cos(\theta)\sin(-\alpha)+\sin(\theta)\cos(-\alpha))\ = \cos(\theta - \alpha) + i \sin(\theta - \alpha)$$

and because we know that the real part has to equal the real part and the complex has to equal the complex we get two equations, which simplify using $\sin(-\theta) = -\sin(\theta)$ and $\cos(-\theta) = \cos(\theta)$ : \begin{align} \cos(\theta)\cos(-\alpha)-\sin(\theta)\sin(-\alpha) &= \cos(\theta + \alpha) &\\ \cos(\theta)\cos(\alpha)+\sin(\theta)\sin(\alpha)&= &\\ \\ \cos(\theta)\sin(-\alpha)+\sin(\theta)\cos(-\alpha) &= \sin(\theta + \alpha) &\\ -\cos(\theta)\sin(\alpha)+\sin(\theta)\cos(\alpha) &=\ &\blacksquare \end{align}

Well, if you buy or have access to (through previous learning) $e^{i\theta} = \cos \theta + i \sin \theta$, then setting $\theta = a - b$ one has

$e^{i(a - b)} = \cos(a - b) + i\sin(a - b). \tag{1}$

But also,

$e^{i(a - b)} = e^{ia}e^{-ib} = (\cos a + i \sin a)(\cos b - i\sin b). \tag{2}$

The imaginary part of the product on the right of (20) is $\sin a \cos b - \cos a \sin b$; the imaginary part of $e^{(a - b)}$ is $\sin (a - b)$. Bringing these imaginary parts together (that is, equating them) yields

$\sin(a - b) = \sin a \cos b - \cos a \sin b. \tag{3}$

as was to be shown. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

With vectors there is a geometric formula for the dot product

$$\vec{u}\cdot\vec{v}=\left\|\vec{u}\right\|\left\|\vec{v}\right\|\cos\theta$$

where $\theta$ is the angle between the two vectors. Take $\vec{u}=\langle\cos(a),\sin(a)\rangle$ and $\vec{v}=\langle-\sin(b),\cos(b)\rangle$ and this formula gives

$$\sin(a)\cos(b)-\cos(a)\sin(b)=\cos\theta$$

where the dot product has been computed using the vector components. Vector $\vec{u}$ points $a$ radians counterclockwise from the positive $x$-axis. The other vector, $\vec{v}$, can be constructed from taking the vector $\langle\cos(b),\sin(b)\rangle$, reflecting it about $y=x$ to get $\langle\sin(b),\cos(b)\rangle$, and then reflecting about $y=0$ to get $\langle-\sin(b),\cos(b)\rangle$. Considering how these reflections affect angles from the positive $x$-axis, this makes $\vec{v}$'s angle be $\pi-\left(\frac{\pi}{2}-b\right)$, which is $\frac{\pi}{2}+b$.

So the value of $\theta$ is $a-\left(\frac{\pi}{2}+b\right)$, and \begin{align} \sin(a)\cos(b)-\cos(a)\sin(b) &=\cos\left(a-\left(\frac{\pi}{2}+b\right)\right)\\ \sin(a)\cos(b)-\cos(a)\sin(b) &=\cos\left(a-b-\frac{\pi}{2}\right)\\ \sin(a)\cos(b)-\cos(a)\sin(b) &=\sin(a-b) \end{align}

$e^{x+y} = e^x e^y$

That is, $e^{i(a+b)} = e^{ia}e^{ib}$

Now, apply the Euler formula to both sides.

• In fact, pretty much the whole litany of (circular) trigonometric identities can be extracted from Euler's Identity. – colormegone Mar 9 '14 at 7:32

$\sin, \cos$ have power series expansions, hence so does $f(t) = \sin(t-b)-( \sin t \cos b - \cos t \sin b)$.

Then $f'(t) = \cos (t-b) -(\cos t \cos b + \sin t \sin b)$, $f''(t) = -f(t)$, etc.

Note that $f(0)=0, f'(0) = 0, f''(0),...$, hence $f(t) = 0$ for all $t$.

$$e^{i\alpha}=\cos \alpha + i\sin \alpha$$

$$e^{i\theta}=\cos \theta + i\sin \theta$$

\begin{align}e^{i(\alpha+\theta)}&=e^{i\alpha}\cdot e^{i\theta} \text{ (law of exponents)}\\ \cos(\alpha+ \theta) + i\sin (\alpha+\theta)&=(\cos\alpha+i\sin\alpha)(\cos \theta + i\sin \theta) \end{align}

Multiply, and compare $\Re(z)$ and $\Im(z)$ of both sides.

From $$e^{ix} = \cos{x} + i\sin{x}$$ we can derive that

$$\sin{x} = \frac{e^{ix} - e^{-ix}}{2i}$$

and

$$\cos{x} = \frac{e^{ix} + e^{-ix}}{2}$$

Hence, we have

$$\sin{a}\cos{b} - \sin{b}\cos{a} = \frac{(e^{ia} - e^{-ia})(e^{ib} + e^{-ib})}{4i} - \frac{(e^{ib} - e^{-ib})(e^{ia} + e^{-ia})}{4i}\\ = \frac{e^{i(a+b)} - e^{i(-a+b)} + e^{i(a-b)} - e^{i(-a-b)} - e^{i(a+b)} + e^{i(a-b)} - e^{i(-a+b)} + e^{i(-a-b)}}{4i}$$ With some cancellations on the numerator, this becomes $$\sin{a}\cos{b} - \sin{b}\cos{a} =\frac{e^{i(a-b)} + e^{-i(a-b)}}{2i}$$ but since $$\sin{x} = \frac{e^{ix} - e^{-ix}}{2i}$$ we conclude $$\sin{a}\cos{b} - \sin{b}\cos{a} = \sin{(a-b)}$$

I searched for an alternate way to deriving sum of angles relations, and by congruence came up with this alternate derivation. Perhaps it is acceptable.

Take a coordinate pair say $(x,y)$ with angle $A$. We have the definitions $\sin(A)= y/(x^2+y^2)^.5$, and $\cos(A)= x/(x^2+y^2)^.5$.

Now rotate, counterclockwise, the coordinate pair $(x,y)$ by an angle $B$ such that
$\angle A + \angle B = \angle C$

After rotation $(x,y)$ becomes $(x\cos B-y\sin B, x\sin B+y\cos B)$. (I'm leaving out a derivation for rotation since it is everywhere, See Rotate the graph of a function?). Since the angle of the rotated pair $(x,y)$ is $\angle C$ the following equations follow:

$$\cos(C)=\frac{x\cos B-y\sin B}{((x\cos B-y\sin B)^2+(x\sin B+y\cos B)^2)^.5}$$

Notice that:

$$(x\cos B-y\sin B)^2+(x\sin B+y\cos B)^2)= (x^2+y^2)$$

Hence,

$$\cos(C)=\frac{x\cos B-y\sin B}{(x^2+y^2)^.5}$$

And

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

Other Relations-------------- $$\sin(C)=\frac{x\sin B+y\cos B}{(x^2+y^2)^.5} = {\cos(A)\sin(B)}+{\sin(A)\cos(B)}$$

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

We also have $$A\tan(y/x)= \angle A$$
and $$A\tan\frac{x\sin B+y\cos B}{x\cos B-y\sin B}= \angle C$$

$$\angle B +A\tan(y/x)= A\tan\frac{x\sin B+y\cos B}{x\cos B-y\sin B}$$

$$\angle B +A\tan(y/x)= A\tan\frac{\tan B+y/x}{1-(y/x)\tan B}$$

Or, for example, by setting $Tan B = u$ and $y/x = v$

$$A\tan(u) +A\tan(v)= A\tan\frac{u+v}{1-uv}$$

For the tangent is straight forward.

$$tan(A + B) = \frac{x\sin B+y\cos B}{x\cos B-y\sin B}$$

$$tan(A + B) = (tan B + tan A) / (1 − tan A *tan B)$$