Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Express sin10+sin20+sin30+sin40+sin50+sin60+sin70+sin80cos5cos10cos20\frac{\sin 10+\sin 20+\sin 30+\sin 40+\sin 50+\sin 60+\sin 70+\sin 80}{\cos 5 \cos 10 \cos 20} without using trigonometric functions.

Solution

Solution:

We will use the identities cosa+cosb=2cosa+b2cosab2\cos a + \cos b = 2 \cos \frac{a+b}{2} \cos \frac{a-b}{2} and sina+sinb=2sina+b2cosab2\sin a + \sin b = 2 \sin \frac{a+b}{2} \cos \frac{a-b}{2}.

The numerator is
(sin10+sin80)+(sin20+sin70)+(sin30+sin60)+(sin40+sin50)(\sin 10 + \sin 80) + (\sin 20 + \sin 70) + (\sin 30 + \sin 60) + (\sin 40 + \sin 50)

=2sin45(cos35+cos25+cos15+cos5)= 2 \sin 45 (\cos 35 + \cos 25 + \cos 15 + \cos 5)

=2sin45((cos35+cos5)+(cos25+cos15))= 2 \sin 45 ((\cos 35 + \cos 5) + (\cos 25 + \cos 15))

=4sin45cos20(cos15+cos5)= 4 \sin 45 \cos 20 (\cos 15 + \cos 5)

=8sin45cos20cos10cos5= 8 \sin 45 \cos 20 \cos 10 \cos 5

So the fraction equals 8sin45=428 \sin 45 = 4 \sqrt{2}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.