Which identity to use?

Geometry Level 5

F = r = 0 ( ( 1 ) r cos 3 3 r 3 r ) \large F=\displaystyle\sum_{r=0}^{\infty}\left((-1)^r\dfrac{\cos^3 3^r}{3^r}\right)

If F F can be written in the form a b cos c d \dfrac ab\cos^c d , where a , b , c a,b,c and d d are positive integers with a , b a,b coprime, find a + b + c + d a+b+c+d .

Clarification : All angles are measured in radians.


The answer is 9.

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1 solution

Rishabh Jain
Jun 19, 2016

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

Use cos 3 x = 4 cos 3 x 3 cos x \cos 3x=4\cos^3x-3\cos x and rearrange for cos 3 x \cos^3 x and substitute in F \mathfrak F :

F = cos 3 1 1 cos 3 3 3 + cos 3 9 9 cos 3 27 27 \mathfrak F=\small{\dfrac{\cos^{3}{1}}{1}-\dfrac{\cos^{3}{3}}{3}+\dfrac{\cos^{3}{9}}{9}-\dfrac{\cos^{3}{27}}{27}\cdots} = 1 4 ( cos 3 + 3 cos 1 1 cos 9 + 3 cos 3 3 + cos 27 + 3 cos 9 9 ) \small{=\dfrac 14\left(\dfrac{\cos 3+3\cos 1}{1}-\dfrac{\cos 9+3\cos 3}{3}+\dfrac{\cos 27+3\cos9}{9}-\cdots\right)}

= 1 4 ( ( cos 3 + 3 cos 1 ) ( cos 9 3 + cos 3 ) + ( cos 27 9 + cos 9 3 ) ) \small{=\dfrac 14\left((\cancel{\color{#20A900}{\cos 3}}+3\cos 1)-(\cancel{\color{#D61F06}{\dfrac{\cos 9}3}}+\cancel{\color{#20A900}{\cos 3}})+(\cancel{\color{#3D99F6}{\dfrac{\cos 27}{9}}}+\cancel{\color{#D61F06}{\dfrac{\cos 9}3}})-\cdots\right)}

= 3 cos 1 4 \large =\dfrac{3\cos 1}4

a = 3 , b = 4 , c = d = 1 a=3,b=4,c=d=1

a + b + c + d = 9 \large a+b+c+d=\boxed{9}

Same approach bro

Aakash Khandelwal - 4 years, 12 months ago

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Yep.. Great

Rishabh Jain - 4 years, 12 months ago

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What was your rank in JEE

Aakash Khandelwal - 4 years, 12 months ago

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