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Find the sum of all possible values of θ ∈ [ − π , π ] \theta \in [-\pi, \pi] such that sin ⁡ 3 ( 1 2 θ ) + sin ⁡ 3 ( 1 3 θ ) + sin ⁡ ( 1 6 θ ) + sin ⁡ ( 1 12 θ ) = 0 \sin^3\left(\dfrac{1}{2}\theta\right) + \sin^3\left(\dfrac{1}{3}\theta\right) + \sin\left(\dfrac{1}{6}\theta\right) + \sin\left(\dfrac{1}{12}\theta\right) = 0


The answer is 0.

Akshaj Kadaveru by Akshaj Kadaveru , 20, USA

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2 solutions

Michael Tang
Dec 25, 2013

Let

f ( x ) = sin ⁡ 3 ( 1 2 x ) + sin ⁡ 3 ( 1 3 x ) + sin ⁡ ( 1 6 x ) + sin ⁡ ( 1 12 x ) . f(x) = \sin^3\left(\dfrac12 x\right) + \sin^3\left(\dfrac13 x\right) + \sin\left(\dfrac16x\right) + \sin\left(\dfrac1{12}x\right).

We want to find the sum of the roots of f f (in the specified range). If f ( θ ) = 0 , f(\theta) = 0, where θ ∈ [ − π , π ] , \theta \in [-\pi, \pi], then

f ( − θ ) = sin ⁡ 3 ( − 1 2 θ ) + sin ⁡ 3 ( − 1 3 θ ) + sin ⁡ ( − 1 6 θ ) + sin ⁡ ( − 1 12 θ ) = − sin ⁡ 3 ( 1 2 θ ) − sin ⁡ 3 ( 1 3 θ ) − sin ⁡ ( 1 6 θ ) − sin ⁡ ( 1 12 θ ) = − f ( θ ) \begin{aligned} f(-\theta) &= \sin^3\left(-\dfrac12 \theta\right) + \sin^3\left(-\dfrac13 \theta\right) + \sin\left(-\dfrac16\theta\right) + \sin\left(-\dfrac1{12}\theta\right)\\ &= -\sin^3\left(\dfrac12 \theta\right) - \sin^3\left(\dfrac13 \theta\right) - \sin\left(\dfrac16\theta\right) - \sin\left(\dfrac1{12}\theta\right) \\ &= -f(\theta)\end{aligned}

so f ( − θ ) = 0 f(-\theta) = 0 as well. We can group each root with its negative, so the sum of all such θ \theta can only be 0 . \boxed{0}.

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Finn Hulse
Apr 7, 2014

It's pretty obvious that for any solution of the equation, its negative will also satisfy. Thus, adding all solutions is simply 0 \boxed{0} . Akshaj, why do you not respond to my questions?

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