# Reasoning that $\sin2x=2 \sin x \cos x$

In mathcounts teacher told us to use the formula $\sin2x=2 \sin x \cos x$. What's the math behind this formula that made it true? Can someone explain? $$\color{red}{\sin 2x}=\color{blue}{2\sin x}\cos x$$

• Wow this makes much more sense!! Thanks man! – Commander Shepard Aug 5 '13 at 14:48
• Sorry i don't get it... – David Aug 5 '13 at 15:35
• @David There are two ways to calculate the red thing: 1) from the right triangle with hypotenuse $1$ (this gives $\sin 2x$) and 2) from the complementary right triangle with blue hypotenuse, equal to $2\sin x$. – Start wearing purple Aug 5 '13 at 15:37
• I can't believe I've never seen this before. Amazing! – Steven Stadnicki Aug 5 '13 at 21:19
• Cool diagram. Quite a simple and elegant geometric demonstration. – Pratyush Sarkar Aug 7 '13 at 23:32

$$\sin(2x)=\mathrm{Im}(e^{2ix})=\mathrm{Im}(e^{ix}e^{ix})=\mathrm{Im}((\cos x+i\sin x)(\cos x+i\sin x))=2\sin x\cos x$$

• $e^{i\phi} = cos \phi + i sin \phi?$ This is Euler's formula, the greatest formula in the whole math! You must always use it to prove trigonometric relationships and integral calculus as it gives simplest and most beatiful solutions and teaches you the imaginary unit! – Val Aug 5 '13 at 19:34
• And for a simple trig problem like this is is extravagant overkill. – marty cohen Aug 7 '13 at 23:13
• @martycohen Not really. It gets the result quickly without any need for geometric cleverness. – Potato Aug 8 '13 at 0:40
• +1: This method also gives all of the double-angle (and triple-angle, etc.) formulas. – Neil G Aug 14 '13 at 22:28
• If we think of usual definition of $\sin x, \cos x$ (i.e. as coordinates of a point revolving on a circle of unit radius), then it is impossible to derive the Euler's formula without the use of addition rules like $\sin(a + b) = \sin a \cos b + \cos a \sin b$. So it becomes circular reasoning. On the other hand if we use the infinite series for $\sin x, \cos x$ as their definition then this is not circular. But I still believe OP did not have infinite series definition in mind. – Paramanand Singh Nov 10 '13 at 11:42

Assuming that you already know the sum of angles formula, this is pretty easy to get:

$\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$, so

$\sin(2x) = \sin(x + x) = \sin(x)\cos(x) + \cos(x)\sin(x) = 2\sin(x)\cos(x)$.

Edit: If you don't want to take the sum of angles formula for sines as a given, this page on Wikipedia can explain it better than I can, especially since it has a diagram.

• I like my picture-proof slightly better than Wikipedia's. :) – Blue Aug 5 '13 at 15:12
• @Blue those pictures are fantastic, thank you. These deserve more exposure. – littleO Aug 7 '13 at 23:28

A rotation matrix is clearly a linear transformation, so by rotating the elementary basis vectors, a two-dimensional rotation matrix for angle $x$ is \begin{align} \begin{bmatrix} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix}. \end{align}

Squaring this matrix is equivalent to applying the rotation twice, so \begin{align} \begin{bmatrix} \cos(2x) & -\sin(2x) \\ \sin(2x) & \cos(2x) \end{bmatrix} &= \begin{bmatrix} \cos^2(x) - \sin^2(x) & -2\sin(x)\cos(x) \\ 2\sin(x)\cos(x) & \cos^2(x) - \sin^2(x) \end{bmatrix}, \end{align} which implies that \begin{align} \cos(2x) &= \cos^2(x) - \sin^2(x) \\ \sin(2x) &= 2\sin(x)\cos(x). \end{align}

Using rotation matrices with different angles gives some of the double-angle (or triple-angle, etc.) identities.

Geometrically, the area of an isosceles triangle with apex $2\alpha$ and leg length $1$ can be computed with one of the legs as base, then the height is $\sin 2\alpha$. So th earea is $\frac12\sin 2\alpha$.

Alternatively, the base has length $2\sin \alpha$ and the corresponding height is $\cos \alpha$, thus the area is $\frac12\cdot2\sin\alpha\cos\alpha$.

Equating both, you get $\sin2\alpha=2\sin\alpha\cos\alpha$.

• @leonbloy Ooops, thanks - edited – Hagen von Eitzen Aug 10 '13 at 9:58

Hint: Find the area of the rectangular ABCD from the triangle AFD and normal way: All of those weird trigonometric identities make sense if you express them as exponentials.

$$\sin 2x = 2\sin x \cos x.$$

Using $e^{xi} = \cos x + i\sin x$ and expressing $\cos$ and $\sin$ through this...

\begin{align} &\frac{e^{2xi} - e^{-2xi}}{2i} = 2 \frac{e^{xi} - e^{-xi}}{2i} \frac{e^{xi} + e^{-xi}}{2},\\ &\frac{e^{2xi} - e^{-2xi}}{2i} = \frac{(e^{xi} - e^{-xi})(e^{xi} + e^{-xi})}{2i},\\ &\frac{e^{2xi} - e^{-2xi}}{2i} = \frac{e^{2xi} - e^{-2xi}}{2i}. \end{align}

Of course, the geometric interpetation makes more sense here.

Let's start with the power series definitions: $$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots,$$ $$\cos x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots.$$ If you expand the product $$\sin x \cos x= \left( x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots \right) \left( 1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots \right),$$ and collect terms, you will find that only odd powers of $x$ appear, and the coefficient of $x^{2n+1}$ is $$(-1)^n \left(\frac{1}{(2n+1)!0!}+\frac{1}{(2n-1)! 2!}+ \dots +\frac{1}{1!(2n)!} \right).$$ You can prove (using induction for example) that the last expression equals $$(-1)^n \frac{2^{2n}}{(2n+1)!}=(-1)^n \frac{1}{2} \frac{2^{2n+1}}{(2n+1)!}.$$ Thus we have $$\sin x \cos x=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!},$$ which is precisely $\frac{1}{2} \sin 2x$.