# Using the unit circle to prove the double angle formulas for sine and cosine?

How do you use the unit circle to prove the double angle formulas for sine and cosine?

• I'd suggest deriving the formula for sums of angles. – user21436 Apr 25 '12 at 13:00
• Use complex numbers. – The Chaz 2.0 Apr 25 '12 at 13:01
• I once saw a direct geometric proof of $\cos\,u=2\cos^2 u-1$, but I don't remember where I saw it... – J. M. isn't a mathematician Apr 25 '12 at 13:36
• @J.M. Here is one geometric proof of $\cos 2\theta=1-2\sin^2\theta$. – David Mitra Apr 25 '12 at 13:56

Look at this figure:

• Ingenious -- but the unit circle is mostly decoration here, isn't it? – hmakholm left over Monica Apr 25 '12 at 15:22
• @HenningMakholm I’d agree with you. If you wanted to make more use of the unit circle, you could draw the diagram with angle $2\alpha$ beginning on the $x$-axis so that $\cos(2\alpha)$ and $2\sin^2\alpha$ ran parallel to the $x$-axis and so that $\sin(2\alpha)$ ran parallel to the $y$-axis. Or, you could use the the above diagram and the just-described diagram to make use of the unit circle entirely. – gen-ℤ ready to perish Jul 26 '17 at 19:28
• The picture seems to show the particular example when $\alpha = \pi/6$. A similar set up clearly works when $\alpha \le \pi/4$. It's not all clear from the answer how to extend this for larger $\alpha$. Hence the -1. – Aaron Dall May 24 at 18:43
• @AaronDall: This figure works for arbitrary $0<\alpha<{\pi\over4}$. When ${\pi\over4}<\alpha<{\pi\over2}$ the value of $\cos(2\alpha)$ is negative, and another figure is needed. – Christian Blatter May 25 at 8:10

Here's one possibility. Say we want to find $\sin 2\theta$ and $\cos 2\theta$. Draw the unit circle in an ordinary $x$-$y$ coordinarte system, and also introduce a new coordinate system $x'$-$y'$ that has been turned $\theta$ clockwise around the origin. It is important that the unit circle in the $xy$ system and in the $x'y'$ system is the same:

The relation between the two coordinate systems is $$x' = x\cos\theta - y\sin\theta \qquad y'=x\sin\theta + y\cos\theta$$

The point $P$ on the diagram has coordinates $(x,y)=(\cos\theta,\sin\theta)$ in the $xy$-system, but in the $x'y'$ system is is $2\theta$ above the $x'$-axis and so its coordinates there must be $(x',y')=(\cos2\theta, \sin2\theta)$. Substituting this into the known relation between the coordinate systems yields: $$\cos2\theta = (\cos \theta)^2 - (\sin\theta)^2 \qquad \sin2\theta = \cos(\theta)\sin(\theta) + \sin(\theta)\cos(\theta)$$

• How did you produce your diagram? Did you draw it with pen and then scan it? – MJD Apr 25 '12 at 14:41
• I drew it with a pen, photographed it with a digital camera, and then cleaned it up (for background and contrast) in Gimp. – hmakholm left over Monica Apr 25 '12 at 14:43
• How did you get the relations x′=xcosθ−ysinθ, y′=xsinθ+ycosθ? – Ralph May 29 '17 at 2:21

This is essentially Christian Blatter's proof, with some minor differences, but I like the area interpretation that this one employs, and the historical connection. It also explains a bit more the connection of Christian Blatter's proof with the circle. This version gives the double-angle formula for $\sin$ only.

A right triangle with hypotenuse $1$ and angle $\theta$ has area $\frac{1}{2}\cos\theta\sin\theta.$ Four such triangles together have area $2\cos\theta\sin\theta.$ Arrange the four right triangles to form a kite-shaped figure.

The two diagonals of the kite-shaped figure (represented by solid lines) are perpendicular, and the area of the figure equals half the product of their lengths. But one diagonal has length $1$ while the other has length $2\sin2\theta$. The double-angle formula for $\sin$ follows.

My interest in this proof is partly historical. Inscribe a regular $n$-gon in the unit circle. Let $\ell_n$ be the length of a side of the $n$-gon. This $n$-gon can be broken into $n$ isosceles triangles with side lengths $1,$ $1,$ $\ell_n.$

Now form a regular $2n$-gon from $n$ kite-shaped figures.

The diagonals of these kite-shaped figures are $1$ and $\ell_n$. The area of this $2n$-gon is therefore $$A_{2n}=\frac{n\ell_n}{2}.$$ By repeatedly doubling the number of sides, $A_{2n}$ becomes an increasingly accurate approximation of $\pi.$ This is the method for estimating $\pi$ devised by Liu Hui around 263 AD. He started with $n=6$ $(\ell_6=1)$ and applied five doublings. In doing this, Liu Hui needed to compute $\ell_{2n}$ from $\ell_n.$ This can be done using the formula $$\ell_{2n}^2=2-\sqrt{4-\ell_n^2},$$ which is essentially the half-angle formula (since $\ell_{2n}=2\sin\theta$ and $\ell_n=2\sin2\theta$), but which Liu Hui derived using the Gougu Theoerem (Pythaogorean Theorem). Chinese geometry of that era apparently did not employ the notion of angle, so the connection with the double-angle and half-angle formulas is ahistorical. Obviously the modern algebraic notation is also ahistorical.

Compute lengths of the thick line segments in the figure below in two different ways.