Trigonometric Sum Solution
Solve
\[2 \sin \theta (\sin 2 \theta + \sin 4 \theta + \sin 6 \theta + \dots + \sin 14 \theta) = \cos \theta - \frac{1}{2}\]where $ 0^\circ \le \theta \le 24^\circ $. Enter all the possible solutions, separated by commas.
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- $\frac{a}{b}$
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Solution
Expanding, we get
\[2 \sin \theta \sin 2 \theta + 2 \sin \theta \sin 4 \theta + 2 \sin \theta \sin 6 \theta + \dots + 2 \sin \theta \sin 14 \theta = \cos \theta - \frac{1}{2}.\]Using the product-to-sum formula, we can write the left-hand side as
\begin{align*}
&2 \sin \theta \sin 2 \theta + 2 \sin \theta \sin 4 \theta + 2 \sin \theta \sin 6 \theta + \dots + 2 \sin \theta \sin 14 \theta \\
&= (\cos \theta - \cos 3 \theta) + (\cos 3 \theta - \cos 5 \theta) + (\cos 5 \theta - \cos 7 \theta) + \dots + (\cos 13 \theta - \cos 15 \theta) \\
&= \cos \theta - \cos 15 \theta..\end{align*}Hence, $ \cos 15 \theta = \frac{1}{2} $.
Since $ 0^\circ \le \theta \le 24^\circ, $ $ 0^\circ \le 15 \theta \le 360^\circ $. Thus, $ 15 \theta = 60^\circ $ or $ 15 \theta = 300^\circ, $ which leads to the solutions $ \boxed{4^\circ, 20^\circ} $.
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