Inscribed Triangles and the Law of Sines

The Law of Sines follows from the following fact:  

Theorem: Any side of a triangle divided by the sine of the opposite angle gives the diameter of the circumscribing circle.

In other words, \( \frac{a}{\sin \alpha}, \frac{b}{\sin \beta}\), and \(\frac{c}{\sin \gamma} \) are all equal because each represents the same diameter.

The Law of Sines can be proven by using the fact that the area of a triangle is half the product of any two sides and the sine of the included angle. But such a proof gives no hint of the geometric interpretation of \( \frac{a}{\sin \alpha}\).

Here's a geometric argument for the theorem.

Case 1: First, we notice that if we have a right triangle with hypotenuse \(c\), then \(\sin\alpha =\frac{a}{c}\), \(\sin\beta =\frac{b}{c}\), and \(\sin\gamma =\sin 90^{\circ}=1\), so all three ratios  \( \frac{a}{\sin \alpha}, \frac{b}{\sin \beta}\), and \(\frac{c}{\sin \gamma} \)  are equal to \(c\), the hypotenuse. And because an inscribed right angle subtends an arc of \(180^{\circ}\), the hypotenuse coincides with a diameter. Thus the theorem is true for all angles in any right triangle.

Case 2: Now suppose that \(\Delta ABC\) is a triangle with \(\alpha =\angle CAB\) an acute angle. Draw the diameter through \(B\) to the point \(D\), and draw the segment from \(D\) to \(C\).

\(\Delta DBC\) is a right triangle inscribed in the same circle and shares the side of length \(a\) with \(\Delta ABC\). The angle at \(D\) subtends the same arc as the angle at \(A\), so the angles are congruent. By Case 1, \(\frac{a}{\sin\alpha}\) is the diameter of the circumscribing circle. Thus the theorem holds for any acute angle in any triangle.

Case 3: Now suppose that  \(\Delta ABC\) is a triangle with \(\alpha =\angle CAB\) an obtuse angle. Choose \(D\) so it lies on the arc of the circle subtended by \(\angle CAB\). Then \(\Delta DBC\) is inscribed in the same circle as \(\Delta ABC\) and shares the side of length \(a\).

Because \(\angle CAB\) and \(\angle CDB\) subtend arcs that sum to \(360^{\circ}\), they are supplementary angles. In particular, \(\angle CDB\) is acute, so the diameter of the circumscribing circle is \(\frac{a}{\sin(180^{\circ}-\alpha)}\). And because \(\sin(180^{\circ}-\alpha) = \sin\alpha\), we conclude that the diameter of the circumscribing circle is  \( \frac{a}{\sin \alpha}\).

Thus, whether \(\alpha\) is a right angle, acute angle, or obtuse angle, the ratio  \( \frac{a}{\sin \alpha}\) gives the diameter of the circumscribing circle.

Cosine of 72 degrees (and constructing a regular pentagon)

By using (say) DeMoivre's theorem, we have that \( \left( \cos\frac{2 \pi}{5}+ i \sin\frac{2\pi}{5} \right)^5=1\)
Expanding the left side as the fifth power of a binomial, equating the imaginary parts on both sides of the equation, and then replacing  \( \sin^2 72^\circ\) with  \(1- \cos^2 72^\circ \)

 \( i\sin 72^\circ \left(5 \cos^4 72^\circ+10 \cos^2 72^\circ i^2 \sin ^2 72 ^\circ+ i^4 \sin^4 72^\circ \right)=0i\)

 \(16 \cos^4 72^\circ - 12 \cos^2 72^\circ  + 1=0\)

Solving this quadratic in \( \cos^2 72^\circ \), we get

 \[ \cos^2 72^\circ  = \frac{12 \pm  \sqrt{80} } {32}   \]

so

 \[ \cos^2 72^\circ  = \frac{6 \pm 2 \sqrt{5} } {16}=\frac{\left( \sqrt{5}\pm 1 \right)^2}{4^2}   \]

 \[ \cos 72^\circ  = \pm \frac{\sqrt{5} \pm 1}{4}  \]

where we can choose the correct value of the four possible values by noting that, because 72° is between 45° and 90°,  \( \cos 72^\circ \) must lie between  \(1 / \sqrt{2}\) and 0. Because  \( \cos 72^\circ \) is positive, we choose the "+" before the fraction, and because  \( \cos 72^\circ \) is less than \( 1 / \sqrt{2}\), which in turn is less than \(\frac{\sqrt{5}+1}{4}\), we choose the "-" in the numerator:

\[ \cos 72^\circ = \frac{\sqrt{5}-1}{4} \]

Constructing a regular pentagon

So we can construct \( \cos 72^\circ\). For example, the diagonal of a 1-by-2 rectangle is \(\sqrt{5}\). We could cut off one unit from a segment of length \(\sqrt{5}\), then divide the segment of length \(\sqrt{5}-1\) into four pieces of length \( \frac{\sqrt{5} -1} {4} \). (Or we could construct the appropriate solution to the equation \( 4x^2 +2x -1 = 0 \). See my post on Solving quadratic equations via geometric construction.)

Construct a unit circle centered at O, and construct a radius \(\overline{OA}\).  Construct the point B on \(\overline{OA}\) so that \(\overline{OB}\) has length  \( \cos 72^\circ\). If C is a point on the circle so that \(\overline{BC}\) is perpendicular to \(\overline{OA}\), then \(\angle COA\) is a 72° angle, and both A and C are vertices of a regular pentagon inscribed in the circle.

sin(n) is dense in [-1,1]

Numbers of the form sin(n) get arbitrarily close to every real value from -1 to 1.

Consider the integers mod(2pi).   Equivalently, we can consider the points (cos n, sin n) on the unit circle (a.k.a. using the "wrapping function").   Because we have an infinite set of points in a compact set, there must be an accumulation point. 

For any epsilon > 0, we can find distinct integers m and n such that  (cos m, sin m)  and  (cos n, sin n)  are less than epsilon apart on the circle, so  (cos(m-n), sin(m-n))  is within epsilon of (1,0).   Then (by say the Archimedean principle),  every point on the unit circle is within epsilon of some integer multiple of (m-n) wrapped around the unit circle .   Thus the points of the form (cos k, sin k)  form a dense subset of the unit circle.  In particular, numbers of the form sin(k) form a dense subset of [-1,1].

A slightly different argument for density of sin(n) depends on the following lemma.

Lemma:  If x is irrational, then the additive group Z + xZ is dense in R.

If we accept the lemma, then the additive group Z + (2pi)Z is dense in R, so the points (cos t, sin t) for t in Z + (2pi)Z must be dense in the unit circle.  But (cos (n+m2pi), sin (n+m2pi)) = (cos n , sin n), so (cos n , sin n) is dense in the circle.

To prove the lemma, consider the contrapositive and assume that Z + xZ is discrete.  Then Z + xZ must be cyclic:  Z + xZ = dZ for some real number d.  Because Z is contained in dZ, there must be some integer n so that 1 = nd, hence d = 1/n is rational.  And then for some m, 1+x = m/n, so x is rational.

And the converse of the lemma is also true, because of x = m/n is rational, then Z + xZ =  1/n Z.

Summing 1/n^2

Here's how Euler evidently first evaluated the sum

We know the MacLaurin series for sin x, and hence

On the other hand, the zeros of this function are the nonzero integer multiples of pi, and writing the series in a factored form, we obtain

Expanding the last (infinite) product of binomials and equating its quadratic coefficient with that of the original MacLaurin series, we obtain

Euler's result follows directly.

Solving quadratic equations via geometric construction

We can solve a quadratic equation of the form x² - sx + p = 0, , using the standard construction tools of compass and straightedge.  The method has been attributed to critic Thomas Carlyle.

Construct the circle in the Cartesian plane with center and passing through A(0,1).  By symmetry, the circle also passes through and .

Because the circle has center   and passes through A(0,1), the equation of the circle is

This reduces to

x² - sx + y² - (p + 1)y + p = 0

So the x-intercepts of the circle are the solutions to  x² - sx + p = 0.

Alternate justification:

The segment joining the x-intercepts has a length , hence x₁ + x₂ = s.

intercepts the arc AX₁X₂, and  intercepts the opposite arc, hence the two angles are supplementary.  But is also supplementary with , so is congruent to , which in turn impies that  .  Hence

 ,

which implies that , so x₁ x₂ = p.

Thus (x -  x₁)(x - x₂) = x² - sx + p, and the solutions to x² - sx + p = 0 are x₁ and x₂ .