How do you use the unit circle to prove the double angle formulas for sine and cosine?
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I'd suggest deriving the formula for sums of angles. – Apr 25 '12 at 13:00
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Use complex numbers. – The Chaz 2.0 Apr 25 '12 at 13:01
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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. ain't a mathematician Apr 25 '12 at 13:36
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1@J.M. Here is one geometric proof of $\cos 2\theta=1-2\sin^2\theta$. – David Mitra Apr 25 '12 at 13:56
4 Answers
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Look at this figure:
Christian Blatter
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Ingenious -- but the unit circle is mostly decoration here, isn't it? – hmakholm left over Monica Apr 25 '12 at 15:22
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@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
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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 '20 at 18:43
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@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 '20 at 08:10
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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: $$ \cos2\theta = (\cos \theta)^2 - (\sin\theta)^2 \qquad \sin2\theta = \cos(\theta)\sin(\theta) + \sin(\theta)\cos(\theta)$$
hmakholm left over Monica
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How did you produce your diagram? Did you draw it with pen and then scan it? – MJD Apr 25 '12 at 14:41
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1I 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
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Compute lengths of the thick line segments in the figure below in two different ways.
Will Orrick
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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.
Will Orrick
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