Baltic Way 1995 · Problem 7

Algebra

Prove that $\sin ^{3} 18^{\circ}+\sin ^{2} 18^{\circ}=1 / 8$ .

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Review

Equations and inequalities

Solution

Solution: We have

\begin{aligned} \sin^{3} 18^{\circ} + \sin^{2} 18^{\circ} &= \sin^{2} 18^{\circ} (\sin 18^{\circ} + \sin 90^{\circ}) \\ &= \sin^{2} 18^{\circ} \cdot 2 \sin 54^{\circ} \cos 36^{\circ} \\ &= 2 \sin^{2} 18^{\circ} \cos^{2} 36^{\circ} \\ &= \frac{2 \sin^{2} 18^{\circ} \cos^{2} 18^{\circ} \cos^{2} 36^{\circ}}{\cos^{2} 18^{\circ}} \\ &= \frac{\sin^{2} 36^{\circ} \cos^{2} 36^{\circ}}{2 \cos^{2} 18^{\circ}} \\ &= \frac{\sin^{2} 72^{\circ}}{8 \cos^{2} 18^{\circ}} \\ &= \frac{1}{8}. \end{aligned}

Results from Baltic Way 1995

9 teams

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