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Irish Mathematical Olympiad

Algebra Difficulty 5.5 AIME, harder Prove it Ireland

Suppose xx is a real number and sin⁡3y=−sin⁡x\sin 3y = -\sin x. Determine all possible values of sin⁡y\sin y.

Solution

Solution 1. Note that
sin⁡3θ=sin⁡θcos⁡2θ+cos⁡θsin⁡2θ=sin⁡θ(1−2sin⁡2θ)+cos⁡θ(2sin⁡θcos⁡θ)=3sin⁡θ−4sin⁡3θ. \sin 3\theta = \sin \theta \cos 2\theta + \cos \theta \sin 2\theta = \sin \theta(1 - 2\sin^2 \theta) + \cos \theta (2\sin \theta \cos \theta) = 3 \sin \theta - 4 \sin^3 \theta.
So
3sin⁡y−4sin⁡3y=−sin⁡x=−(3sin⁡z−4sin⁡3z),z=x/3. 3 \sin y - 4 \sin^3 y = - \sin x = -(3 \sin z - 4 \sin^3 z), \quad z = x/3.
Or with t=2sin⁡yt = 2 \sin y, s=2sin⁡zs = 2 \sin z,
3t−t3=−3s+s3,(t+s)(t2−ts+s2−3)=0. 3t - t^3 = -3s + s^3, \quad (t+s)(t^2 - ts + s^2 - 3) = 0.
Thus, either t=−st = -s or
t=s±s2−4(s2−3)2=s±34−s22=sin⁡z±3∣cos⁡z∣, t = \frac{s \pm \sqrt{s^2 - 4(s^2 - 3)}}{2} = \frac{s \pm \sqrt{3\sqrt{4-s^2}}}{2} = \sin z \pm \sqrt{3} |\cos z|,
Since
∣sin⁡z±3∣cos⁡z∣∣≤1+3(sin⁡2z+cos⁡2z)=2, |\sin z \pm \sqrt{3} |\cos z|| \le \sqrt{1+3(\sin^2 z + \cos^2 z)} = 2,
this means that there are three possible values for sin⁡y\sin y, namely
−sin⁡(x3),12(sin⁡(x3)±3cos⁡(x3)). -\sin\left(\frac{x}{3}\right), \quad \frac{1}{2}\left(\sin\left(\frac{x}{3}\right) \pm \sqrt{3}\cos\left(\frac{x}{3}\right)\right).

Solution 2. To solve the equation sin⁡(3y)+sin⁡x=0\sin(3y) + \sin x = 0, we use the trigonometric identity that converts a sum of sines to a product to get
2sin⁡(3y+x2)cos⁡(3y−x2)=0. 2 \sin \left( \frac{3y+x}{2} \right) \cos \left( \frac{3y-x}{2} \right) = 0.
Therefore, sin⁡((3y+x)/2)=0\sin((3y + x)/2) = 0 or cos⁡((3y−x)/2)=0\cos((3y - x)/2) = 0, so
3y+x2=nπor3y−x2=(n+12)πfor some n∈Z. \frac{3y+x}{2} = n\pi \quad \text{or} \quad \frac{3y-x}{2} = \left(n+\frac{1}{2}\right)\pi \quad \text{for some } n \in \mathbb{Z}.
Hence y=13(2nπ−x)y = \frac{1}{3}(2n\pi - x) or y=13((2n+1)π+x)y = \frac{1}{3}((2n + 1)\pi + x), for some n∈Zn \in \mathbb{Z}. So sin⁡y=sin⁡(13(2nπ−x))\sin y = \sin(\frac{1}{3}(2n\pi - x)) or sin⁡y=sin⁡(13((2n+1)π+x))\sin y = \sin(\frac{1}{3}((2n+1)\pi+x)), for some n=0,1,2n = 0, 1, 2. Here we have used the periodicity of sine to restrict the values of nn. Using the identity sin⁡(π−α)=sin⁡α\sin(\pi - \alpha) = \sin \alpha, the latter three solutions can be written as sin⁡y=sin⁡(π−13((2n+1)π−x))=sin⁡(13((2−2n)π−x))\sin y = \sin(\pi - \frac{1}{3}((2n+1)\pi - x)) = \sin(\frac{1}{3}((2 - 2n)\pi - x)), for n=0,1,2n = 0, 1, 2. These give the same as the first three solutions. So the three solutions are sin⁡y=sin⁡(13(2nπ−x))\sin y = \sin(\frac{1}{3}(2n\pi - x)), for n=0,1,2n = 0, 1, 2, and this is the same as above.

−sin⁡(x3),12(sin⁡(x3)±3cos⁡(x3)). -\sin\left(\frac{x}{3}\right), \quad \frac{1}{2}\left(\sin\left(\frac{x}{3}\right) \pm \sqrt{3}\cos\left(\frac{x}{3}\right)\right).

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