Solution to puzzle 111: Trigonometric progression

Complex Arithmetic Solution

We will make use of Euler's formula, which states that, for any real number t,  eit = cos t + i sin t.
It follows that
cos t = ½ (eit + e−it), and
i sin t = ½ (eit − e−it).

Let C = (cos a + cos 2a + ... + cos na), and S = (sin a + sin 2a + ... + sin na).
Then consider P = eia + e2ia + ... + eina = C + iS.

Evaluating P as a geometric series, we have

Recall that P = C + iS.  That is, C is the real part of the above expression for P; S is the imaginary part.

Trigonometric Solution

We will make use of the following product-to-sum and sum-to-product trigonometric identities.

 sin x · sin y = ½ [cos(x − y) − cos(x + y)] (1) sin x · cos y = ½ [sin(x + y) + sin(x − y)] (2) cos x − cos y = −2 sin ½(x + y) · sin ½(x − y) (3) sin x − sin y = 2 cos ½(x + y) · sin ½(x − y) (4)

Let S = (sin a + sin 2a + ... + sin na).

 Then S · sin(a/2) = ½ [cos(a/2) − cos(3a/2) + cos(3a/2) − cos(5a/2) + ... + cos[(2n−1)a/2] − cos[(2n+1)a/2]],  by (1). = ½ [cos(a/2) − cos[(2n+1)a/2]]. = sin[(n+1)a/2] · sin(na/2),  by (3).

Similarly, let C = (cos a + cos 2a + ... + cos na).

 Then C · sin(a/2) = ½ [−sin(a/2) + sin(3a/2) − sin(3a/2) + sin(5a/2) − ... − sin[(2n−1)a/2] + sin[(2n+1)a/2]],  by (2). = ½ [−sin(a/2) + sin[(2n+1)a/2]]. = cos[(n+1)a/2] · sin(na/2),  by (4).

Six tangents

Show that tan (/13) · tan (2/13) · tan (3/13) · tan (4/13) · tan (5/13) · tan (6/13) = .

`Hint  -  Solution`

Center of mass

Pennies are placed around the circumference of a circle as follows.  Using polar coordinates, and with the center of the circle at the pole, place 1 penny with its center of mass vertically above the point (1,0), 2 pennies piled vertically at (1,a), 3 pennies at (1,2a), ... , n pennies at (1,(n−1)a), where a = 2/n radians.  Find the x and y coordinates of the center of mass of the coins.