## Posts Tagged ‘Trigonometry’

### Monday Math 158

August 11, 2014

Find a non-summation expression for the value of the sum $\cos{x}+2\cos{2x}+3\cos{3x}+\cdots+n\cos{nx}$.

### Monday Math 143

November 15, 2010

Find .
Solution:

### Monday Math 141

November 1, 2010

Let us combine the results from the previous two weeks (here and here). We found that for a regular n-gon with unit sides, the diagonals have lengths $d_k=\frac{\sin\frac{k\pi}n}{\sin\frac{\pi}n}$, k=2,3,…,n-2, with d1=dn-1=1 the length of the sides.
Now, then, we consider the product of two diagonal lengths, dkdm. From the above, this is
$d_kd_m=\frac{\sin\frac{k\pi}n\sin\frac{m\pi}n}{\sin^2\frac{\pi}n}$.
We found here that the numerator product is
$\sin\frac{k\pi}n\sin\frac{m\pi}n=\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n$.
Thus,
$\begin{array}{rcl}d_kd_m&=&\frac{\sin\frac{k\pi}n\sin\frac{k\pi}n}{\sin^2\frac{\pi}n}\\&=&\frac1{\sin^2\frac{\pi}n}\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n\\&=&\frac1{\sin\frac{\pi}n}\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n\\d_kd_m&=&\sum_{i=1}^{m}\frac{\sin\frac{(k-m+2i-1)\pi}n}{\sin\frac{\pi}n}\end{array}$.
Now, if km, then km+2i-1 is a positive integer for all i=1,2,…,m; similarly, if k+mn, then the largest value of km+2i-1 in the sum, k+m-1, is then an integer less than n, and thus the term $\frac{\sin\frac{(k-m+2i-1)\pi}n}{\sin\frac{\pi}n}$ is equal to a diagonal (or side) $d_{k-m+2i-1}$ for all i in the sum, so
$d_kd_m=\sum_{i=1}^{m}d_{k-m+2i-1}$, with 2≤mkn-2, k+mn. This is the “diagonal product formula” named by Peter Steinbach, who outlined a proof using the n-gon formed by the complex n-roots of unity (compare this problem, for example).
Now, we can generalize the formula first by exchanging k and m to see that for mk,
$d_kd_m=\sum_{i=1}^{k}d_{m-k+2i-1}$;
we combine to form
$d_kd_m=\sum_{i=1}^{\min(k,m)}d_{|k-m|+2i-1}$,
which gives us the formula for any 2≤kn-2, 2≤mn-2, k+mn.
For k+mn, we see that one or both of k and m must be greater than $\frac{n}2$. However, recall that dk=dnk. Thus, if $d_k>\frac{n}2$, then $d_{n-k}<\frac{n}2$, and similarly for m; thus, we see we can pick the smaller of k and nk; the latter gives the same terms in the sum, just in opposite order, so only the limit matters. Doing the same with m, we thus find the formula for all k and m in the range 2 to n-2:
$d_kd_m=\sum_{i=1}^{\min(k,m,n-k,n-m)}d_{|k-m|+2i-1}$,

Now, let us look at a few examples. The smallest n to give us a diagonal is n=4, the square. For square of unit side, the length of the diagonal is $\sqrt2$, and so the only diagonal product is
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$(\sqrt2)^2=2=1+1$
For the pentagon, the diagonals have length $\phi=\frac{1+\sqrt5}2$, the golden ratio, so the only unique diagonal product, since $d_2^2=d_2d_3=d_3^2$, is
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$\phi^2=1+\phi$,
a classic equation for the golden ratio.
For the hexagon, we have $d_2=d_4=\sqrt{3}$ and $d_3=2$.
Thus we have three unique products
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$(\sqrt3)^2=3=1+2$;
$d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4$, which is
$2(\sqrt3)=\sqrt{3}+\sqrt{3}$;
and
$d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5$, which is
$2^2=4=1+2+1$.

Further, this can be used to prove some interesting relations of the (non-constructible) diagonal lengths of the heptagon, and thus the sides of the heptagonal triangle. Letting the heptagon have unit side, labelling the shorter and longer diagonal lengths as $a=\frac{\sin\frac{2\pi}7}{\sin\frac{\pi}7}$ and $b=\frac{\sin\frac{3\pi}7}{\sin\frac{\pi}7}$ respectively, then our diagonal product formula tells us (with $d_1=d_6=1$, $d_2=d_5=a$, and $d_3=d_4=b$):
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, and thus
$a^2=1+b$;
$d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4$, and thus
$ab=a+b$;
and
$d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5$, which is
$b^2=1+a+b$.
Via simple algebra on these equations, we can find formula for the quotients of these numbers; dividing $ab=a+b$ by a or b and solving for the quotient left, we see
$\frac{b}{a}=b-1$ and $\frac{a}{b}=a-1$;
and dividing by the product ab gives
$\frac{1}{a}+\frac{1}{b}=1$,
while further algebra with these formulas lets us find the reciprocals of a and b as linear combinations of a, b, and 1:
$\begin{array}{rcl}b^2&=&1+a+b\\b&=&\frac{1}b+\frac{a}{b}+1\\b&=&\frac{1}b+a-1+1\\b&=&\frac{1}b+a\\\frac{1}b&=&b-a\end{array}$
and
$\begin{array}{rcl}a^2&=&1+b\\a&=&\frac{1}a+\frac{b}{a}\\a&=&\frac{1}a+b-1\\\frac{1}a&=&a-b+1\end{array}$.

These relations appear, for example, in the substitution rules for Danzer’s 7-fold quasiperiodic tiling and Maloney’s 7-fold quasiperiodic tiling

### Monday Math 140

October 25, 2010

What are the lengths of the diagonals of a regular n-sided polygon with sides of unit length?
(more…)

### Monday Math 139

October 18, 2010

Prove that for positive integers k, m, and n, that
$\sin\frac{k\pi}n\sin\frac{m\pi}n=\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n$.
Solution:

### Monday Math 119

May 17, 2010

Very few of the calculus textbooks I have used give a rigorous derivation of the derivatives of sine and cosine, instead using little more than the graphs of the functions as justification. Here, I will demonstrate a rigorous derivation, starting with a clear derivation of the limit .

[Click on figure for full size image]

Now, the area of the triangle ▵OAB is . Similarly, the area of the circular sector between OA and OB is . Lastly, the right triangle ▵OAD has area .
Comparing these areas, we have inequality , which means

Multiplying this by the positive quantity , we get
,
which, using , becomes
.
Therefore, the inverses obey the inequality
,
and since , by the squeeze theorem, we see
; and since  is an even function, this must also be the left-hand limit, and so we have
.

Now, consider . We can find this limit using the above limit and a little trigonometry:
.

Using these two limits and the addition formulas for sine and cosine, we compute the derivatives from the definition .
First,
,
and second,
.

### Monday Math 40

October 6, 2008

Suppose we have three congruent, mutually (externally) tangent circles of radius r. If we circumscribe them with a larger circle of radius R, so that they are all internally tangent to the large circle, what is R in terms of r? What if, instead of three smaller circles, we have a symmetric ring of n≥2 circles inscribed inside the larger circle?

Let point O be the center of the large circle, point A be the center of one of the smaller circles, and B be one of that circle’s points of tangency with one of the other smaller circles. Let us denote the angle ∠AOB as θ.

Figure 1: Our construction

From our figure of this, we see that the length of segment OA is Rr, and that the length of AB is r. But the ratio of AB to OA is the sine of the angle θ. Here, we see θ is equal to π/3.
Thus
.

Similarly for the case with n circles, we simply have a different value for
θ: .

Figure 2: The n=4 case

Thus,
.