Solution to puzzle 111: Trigonometric progression

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Complex Arithmetic Solution

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

Let C = (cos a + cos 2a + ... + cos na), and S = (sin a + sin 2a + ... + sin na).
Then consider P = e^ia + e^2ia + ... + e^ina = 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).

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Six tangents

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

Hint  -  Solution

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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.

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Further reading

1. Prosthaphaeresis Formulas

Source: Trigonometric Delights, by Eli Maor. See Chapter 8.

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