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Geometry Level 3

k = 1 3 cos 2 ( ( 2 k 1 ) π 12 ) \sum_{k=1}^3 \cos^2 \left( (2k-1) \dfrac \pi{12} \right)

If the summation above can be expressed as a b \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 5.

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1 solution

Alex G
May 28, 2016

cos 2 ( π 12 ) + cos 2 ( π 4 ) + cos 2 ( 5 π 12 ) \cos^2{(\dfrac{\pi}{12})}+ \cos^2{(\dfrac{\pi}{4})} + \cos^2{(\dfrac{5\pi}{12})}

cos 2 x = 1 + cos 2 x 2 \cos^2 x= \dfrac{1+\cos{2x}}{2}

1 + cos π 6 2 + 1 2 + 1 + cos 5 π 6 2 \dfrac{1+\cos \dfrac{\pi}{6} }{2}+ \dfrac{1}{2}+ \dfrac{1+\cos \dfrac{5\pi}{6} }{2}

1 + 3 2 2 + 1 2 + 1 3 2 2 \dfrac{1+\dfrac{\sqrt{3}}{2}}{2} + \dfrac{1}{2} + \dfrac{1-\dfrac{\sqrt{3}}{2}}{2}

3 2 \dfrac{3}{2}

5 \boxed{5}

very nice solution. Alex G.

Ramiel To-ong - 4 years ago

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