1997 AIME Problem 11

Question

Let

x = \frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle\sum_{n=1}^{44} \sin n^\circ}.

What is the greatest integer that does not exceed

100x?

Answer

Multiply numerator and denominator by

2\sin\frac{1}{2}^\circ

and telescope using product-to-sum identities

Sum-to-product turns the result into

\cot 22.5^\circ,

which the half-angle formula evaluates exactly

Multiply numerator and denominator by

2\sin\frac{1}{2}^\circ.

Since

2\cos n^\circ \sin\frac{1}{2}^\circ

= \sin\left(n + \frac{1}{2}\right)^\circ

- \sin\left(n - \frac{1}{2}\right)^\circ

and

2\sin n^\circ \sin\frac{1}{2}^\circ

= \cos\left(n - \frac{1}{2}\right)^\circ

- \cos\left(n + \frac{1}{2}\right)^\circ,

both sums telescope:

\begin{aligned} x &= \frac{\sin 44.5^\circ - \sin 0.5^\circ}{\cos 0.5^\circ - \cos 44.5^\circ} \\ &= \frac{2\cos 22.5^\circ \sin 22^\circ}{2\sin 22.5^\circ \sin 22^\circ} \\ &= \cot 22.5^\circ, \end{aligned}

using the sum-to-product identities in the last step.

By the half-angle formula,

\cot 22.5^\circ = \frac{1 + \cos 45^\circ}{\sin 45^\circ} = \sqrt{2} + 1.

Since

1.41^2 \lt 2 \lt 1.42^2,

we have

241 \lt 100\sqrt{2} + 100 \lt 242.

Therefore the greatest integer not exceeding

100x

is

241.

Problem 11 in other years