What is the Sine of 18°?

To find the exact values of sine (and cosine) of a given angle, you need to use identities. This particular value is interesting in that it takes a fair bit of work to arrive at this value analytically, and yet the value is expressible as a fairly simple radical. The first thing to notice is that 90 = 5(18). You probably know the half angle identity, but not a fifth angle identity.  And for good reason—the fifth angle identity has multiple solutions; it's a fifth degree polynomial.

I did this calculation manually the first time through—some years ago. It goes a lot easier with a computer algebra system. Today, I'm going to use Maxima to solve the problem.

An identity which comes up in the development of the complex numbers, gives a convenient way of expressing \(\sin {n\theta}\) in terms of combinations of \(\sin \theta\) and \(\cos \theta\).

De Moivre's formula:
\[r^n (\cos \theta + i \sin \theta)^n = r^n (\cos {n\theta} + i \sin {n\theta}).\]
So, if we take \(r=1\) and \(n=5\), we can do a binomial expansion and easily isolate the \(sin \theta\) and \(\cos \theta\). In a Maxima cell, I type

eq: (cos(%theta) + %i*sin(%theta))^5 = cos(5*%theta) + %i*sin(5*%theta);
realpart(eq);
imagpart(eq);

After executing, I obtain the fifth angle identities (if you want to call them that).
\[\mathrm{cos}\left( 5\,\theta\right) =5\,\mathrm{cos}\left( \theta\right) \,{\mathrm{sin}\left( \theta\right) }^{4}-10\,{\mathrm{cos}\left( \theta\right) }^{3}\,{\mathrm{sin}\left( \theta\right) }^{2}+{\mathrm{cos}\left( \theta\right) }^{5}\]
\[\mathrm{sin}\left( 5\,\theta\right) ={\mathrm{sin}\left( \theta\right) }^{5}-10\,{\mathrm{cos}\left( \theta\right) }^{2}\,{\mathrm{sin}\left( \theta\right) }^{3}+5\,{\mathrm{cos}\left( \theta\right) }^{4}\,\mathrm{sin}\left( \theta\right) \]
To eliminate \(\cos^2 \theta\) from the sine identity we use

subst([cos(%theta)=sqrt(1-sin(%theta)^2)],%);
expand(%);

and obtain
\[\mathrm{sin}\left( 5\,\theta\right) =16\,{\mathrm{sin}\left( \theta\right) }^{5}-20\,{\mathrm{sin}\left( \theta\right) }^{3}+5\,\mathrm{sin}\left( \theta\right). \]
Taking \(\theta=18°\), using \(x\) to stand in for \(\sin 18°\), and rearranging, we have
\[0 = 16\,{x}^{5}-20\,{x}^{3}+5\,x - 1.\]
Factoring this one manually was made easier by recognizing \(x = 1\) as a root. After that I tried forming a square free polynomial and found that there was in fact a squared polynomial involved. In general, this helps in a couple ways (potentially):

1. Reduces the degree of the polynomial to be factored.
2. Sometimes produces fully reduced factors as a by-product of forming the square free polynomial.
1. This happens when each repeated factor is of different degree; that is, occurs a different number of times.

(Square free means no factors left that are raised to some power—each factor occurs exactly once.) Doing this manually involves taking the greatest common divisor between a polynomial and it's derivative. But today, let's let Maxima do it.

factor(0=16*x^5-20*x^3+5*x - 1);

\[0=\left( x-1\right) \,{\left( 4\,{x}^{2}+2\,x-1\right) }^{2}.\]

At this point, you really do have to be exceedingly lazy to not continue with the rest of the work manually. It's just a quadratic formula to be applied now (we discard \(x=1\) as an artifact). To use a computer algebra system for the rest is...well...okay, it is Saturday [at original publication], so...

solve(0=4*x^2+2*x-1,[x]);

\[[x=-\frac{\sqrt{5}+1}{4},x=\frac{\sqrt{5}-1}{4}]\]

We reject the negative solution (which is \(\sin {-54°}\) and, interestingly, is the negative of half the Fibonacci number) on the grounds that the range of the sine function on our interval is positive. Hence,

\[\sin 18° = \frac {\sqrt{5} - 1}{4}.\]

Field Measurement of Quadrilaterals Using Only a Tape Measure

Trapezoids

A common nonrectangular area that needs to me measured in the field is a trapezoid. Sometimes shapes are approximated by a trapezoid, if a rectangle is not considered a reasonable approximation or not easily visualized. The well-known formula for the area of a trapezoid is

                        

where b₁ and b₂ are the lengths of the two parallel sides and h is the distance between them (measured perpendicularly to the parallel sides, of course). It is noteworthy that the (b₁+b₂)/2 can be interpreted as the average of the lengths of the parallel sides. This is the length of a line referred to as the median of the trapezoid. In the field, it is often more convenient/faster to measure the median and the height (h, also called the altitude) than to measure both bases and the height. If the location of the median can be “eye-balled” with sufficient precision for the purposes being met, this may increase productivity by decreasing the number of measurements needed. For more information on the trapezoid and the median, see a description here.

General Case Quadrilateral

There are (at least) two approaches to measuring the area quadrilaterals when nothing is known about the internal angles or “parallelness” of nonadjacent lines. The following diagram displays both methods.

We can determine the area of this shape by making 5 measurements and produce an exact picture, or by taking 3 measurements and getting an approximate area. Both are “area by triangles” methods.

Area by Triangles – Heron Method

If we measure the sides AB, BC, CD, DA, and one of the diagonals (either AC or BD), we can use Heron’s theorem to determine the area of the triangles on either side of the chosen diagonal. We can use the law of cosines and law of sines to determine the angles. Alternatively, a basic AutoCAD drawing using temporary construction-line circles would allow you to determine these results without manual calculations. This would give us a full description of the area. This method also generalizes well to an arbitrary number of sides (but requiring a lot of measurements--don't use this method).

Area by Triangles – By Altitude

On the other hand, we could measure the length of a diagonal (say, AC), leave the measuring tape in place (or replace with a string line) and use a second tape measure to measure from the other vertices (B and D) to the diagonal. As long as the accuracy/precision of the result desired is not too high, we can “eye-ball” perpendicular to the diagonal to get the heights of the triangles to an acceptable accuracy. The area calculation is obvious from there. This method involves fewer measurements, minimal extra equipment, and is quicker to calculate, which may make it more efficient for a quadrilateral. It does not generalize to an n-sided figure as easily as the previous method, but could be done with additional string lines used simultaneously.