Returns back to square one

Geometry Level 3

If f ( x ) = sin x cos 2 x + sin 3 x cos 4 x + sin 5 x f(x) = \sin x - \cos^2 x + \sin^3 x - \cos^4 x + \sin^5 x - \ldots , what is 3 f ( π 3 ) 3 f \left ( \frac \pi 3 \right ) ?

4 3 1 4\sqrt { 3 } -1 7 3 1 7\sqrt { 3 } -1 6 3 1 6\sqrt { 3 } -1 5 3 1 5\sqrt { 3 } -1

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2 solutions

Chew-Seong Cheong
Sep 26, 2018

f ( x ) = sin x cos 2 x + sin 3 x cos 4 x + sin 5 x = sin x + sin 3 x + sin 5 x + ( cos 2 x + cos 4 x + cos 6 x + ) f ( π 3 ) = 3 2 + ( 3 2 ) 3 + ( 3 2 ) 5 + ( 1 4 + ( 1 4 ) 2 + ( 1 4 ) 3 + ) = 3 2 ( 1 1 3 4 ) 1 4 ( 1 1 1 4 ) = 2 3 1 3 \begin{aligned} f(x) & = \sin x - \cos^2 x + \sin^3 x - \cos^4 x + \sin^5 x - \cdots \\ & = \sin x + \sin^3 x + \sin^5 x + \cdots - (\cos^2 x + \cos^4 x + \cos^6 x + \cdots ) \\ f\left(\frac \pi 3\right) & = \frac {\sqrt 3}2 + \left(\frac {\sqrt 3}2\right)^3 + \left(\frac {\sqrt 3}2\right)^5 + \cdots - \left(\frac 14 + \left(\frac 14 \right)^2 + \left(\frac 14 \right)^3 + \cdots \right) \\ & = \frac {\sqrt 3}2 \left(\frac 1{1-\frac 34}\right) - \frac 14 \left(\frac 1{1-\frac 14}\right) \\ & = 2\sqrt 3 - \frac 13 \end{aligned}

Therefore, 3 f ( π 3 ) = 6 3 1 3f\left(\dfrac \pi 3\right) = \boxed{6\sqrt 3-1} .

sin x 1 sin 2 x cos 2 x 1 cos 2 x \dfrac {\sin x}{ 1 - \sin^2x} - \dfrac {\cos^2x}{1 - \cos^2x}

3 f ( π 3 ) = 3 ( 3 / 2 1 1 / 4 1 / 4 1 1 / 4 ) = 3 ( 2 3 1 / 3 ) = 6 3 1 3f(\dfrac {π}{3}) = 3 (\dfrac {\sqrt{3}/2}{1 - 1/4} - \dfrac {1/4}{1 - 1/4}) = 3(2\sqrt{3} - 1/3) = 6\sqrt{3} - 1

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