Mysterious angles

Geometry Level 3

1 + sin ⁡ x + cos ⁡ 2 x sin ⁡ 2 x + cos ⁡ x = 3 \large \frac{1+\sin {x}+\cos {2x}}{\sin{2x}+\cos{x}}=\sqrt{3}

x = π X x=\frac \pi X is a solution to the equation above for some positive integer X X . What is X ? X?


Hint: See this note .


The answer is 30.

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Noel Lo by Noel Lo , 24, Singapore

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2 solutions

Noel Lo
Jul 26, 2017

1 + sin ⁡ x + cos ⁡ 2 x sin ⁡ 2 x + cos ⁡ x = 3 \frac{1+\sin{x}+\cos{2x}}{\sin{2x}+\cos{x}}=\sqrt{3}

1 + sin ⁡ x + cos ⁡ 2 x = 3 ( sin ⁡ 2 x + cos ⁡ x ) 1+\sin{x}+\cos{2x}=\sqrt{3}(\sin{2x}+\cos{x})

1 2 ( 1 + sin ⁡ x + cos ⁡ 2 x ) = 3 2 ( sin ⁡ 2 x + cos ⁡ x ) \frac{1}{2}(1+\sin{x}+\cos{2x})=\frac{\sqrt{3}}{2}(\sin{2x}+\cos{x})

1 2 + 1 2 sin ⁡ x + 1 2 cos ⁡ 2 x = 3 2 sin ⁡ 2 x + 3 2 cos ⁡ x \frac{1}{2}+\frac{1}{2}\sin{x}+\frac{1}{2}\cos{2x}=\frac{\sqrt{3}}{2}\sin{2x}+\frac{\sqrt{3}}{2}\cos{x}

1 2 + sin ⁡ π 6 sin ⁡ x + cos ⁡ π 3 cos ⁡ 2 x = sin ⁡ π 3 sin ⁡ 2 x + cos ⁡ π 6 cos ⁡ x \frac{1}{2}+\sin{\frac{\pi}{6}}\sin{x}+\cos{\frac{\pi}{3}}\cos{2x}=\sin{\frac{\pi}{3}}\sin{2x}+\cos{\frac{\pi}{6}}\cos{x}

1 2 + cos ⁡ π 3 cos ⁡ 2 x − sin ⁡ π 3 sin ⁡ 2 x = cos ⁡ π 6 cos ⁡ x − sin ⁡ π 6 sin ⁡ x \frac{1}{2}+\cos{\frac{\pi}{3}}\cos{2x}-\sin{\frac{\pi}{3}}\sin{2x}=\cos{\frac{\pi}{6}}\cos{x}-\sin{\frac{\pi}{6}}\sin{x}

1 2 + cos ⁡ ( π 3 + 2 x ) = cos ⁡ ( π 6 + x ) \frac{1}{2}+\cos{(\frac{\pi}{3}+2x)}=\cos{(\frac{\pi}{6}+x)}

1 2 + cos ⁡ 2 ( π 6 + x ) = cos ⁡ ( π 6 + x ) \frac{1}{2}+\cos{2(\frac{\pi}{6}+x)}=\cos{(\frac{\pi}{6}+x)}

Now we let θ = π 6 + x \theta=\frac{\pi}{6}+x . Note that we choose the special angles of π 3 \frac{\pi}{3} and π 6 \frac{\pi}{6} carefully so that we end up with this result:

1 2 + cos ⁡ ( 2 θ ) = cos ⁡ θ \frac{1}{2}+\cos{(2\theta)}=\cos{\theta}

It then remains to make use of the hint given in the question:

2 sin ⁡ θ ( 1 2 + cos ⁡ ( 2 θ ) ) = 2 sin ⁡ θ cos ⁡ θ 2\sin{\theta}(\frac{1}{2}+\cos{(2\theta)})=2\sin{\theta}\cos{\theta}

sin ⁡ θ + 2 sin ⁡ θ cos ⁡ ( 2 θ ) = sin ⁡ ( 2 θ ) \sin{\theta}+2\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}

sin ⁡ ( 2 − 1 ) θ + 2 sin ⁡ θ cos ⁡ ( 2 θ ) = sin ⁡ ( 2 θ ) \sin{(2-1)\theta}+2\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}

sin ⁡ ( 2 θ ) cos ⁡ θ − sin ⁡ θ cos ⁡ ( 2 θ ) + 2 sin ⁡ θ cos ⁡ ( 2 θ ) = sin ⁡ ( 2 θ ) \sin{(2\theta)}\cos{\theta}-\sin{\theta}\cos{(2\theta)}+2\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}

sin ⁡ ( 2 θ ) cos ⁡ θ + sin ⁡ θ cos ⁡ ( 2 θ ) = sin ⁡ ( 2 θ ) \sin{(2\theta)}\cos{\theta}+\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}

sin ⁡ ( 2 + 1 ) θ = sin ⁡ ( 2 θ ) \sin{(2+1)\theta}=\sin{(2\theta)}

sin ⁡ ( 3 θ ) = sin ⁡ ( 2 θ ) \sin{(3\theta)}=\sin{(2\theta)}

sin ⁡ ( 3 θ ) = sin ⁡ ( π − 2 θ ) \sin{(3\theta)}=\sin{(\pi-2\theta)}

3 θ = π − 2 θ 3\theta=\pi-2\theta

( 3 + 2 ) θ = π (3+2)\theta=\pi

5 θ = π 5\theta=\pi

θ = π 5 \theta=\frac{\pi}{5}

Therefore,

π 6 + x = π 5 \frac{\pi}{6}+x=\frac{\pi}{5}

x = π 5 − π 6 x=\frac{\pi}{5}-\frac{\pi}{6}

x = π 30 x=\frac{\pi}{30}

As required, X = 30 X=\boxed{30}

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Elegant solution! Upvoted.

Rishik Jain - 3 years, 10 months ago

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Thanks for your encouragement!!! (+1)

Noel Lo - 3 years, 10 months ago

very nice. I seem to recall someone, perhaps @Mark Hennings , had an interesting solution to this as well. Mark, if so, would you mind reposting it?

Wesley Zumino - 3 years, 10 months ago

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My technique was to start with the equation 1 2 + cos ⁡ ( 2 ( x + 1 6 π ) ) = cos ⁡ ( x + 1 6 π ) \tfrac12 + \cos\big(2(x+\tfrac16\pi)\big) \; = \; \cos(x + \tfrac16\pi) and turn it into a quadratic equation for cos ⁡ ( x + 1 6 π ) \cos(x + \tfrac16\pi) . This gives two possible values for cos ⁡ ( x + 1 6 π ) \cos(x + \tfrac16\pi) , one of which is 1 4 ( 1 + 5 ) = cos ⁡ 1 5 π \tfrac14(1 + \sqrt{5}) = \cos\tfrac15\pi .

Mark Hennings - 3 years, 10 months ago

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The quadratic equation approach is nice!

Calvin Lin Staff - 3 years, 10 months ago

Please be very careful with the equivalence of equations. sin ⁡ A = sin ⁡ B \sin A = \sin B does not imply that A = B A = B .

This is a very common mistake made when solving trigonometric equations. Can you determine the complete solution set to the problem?

Calvin Lin Staff - 3 years, 10 months ago

Your solution is incomplete. You have only demonstrated that pi/30 can be a solution of x. But in fact, there are infinitely many solutions of x. Only one of which is a unit fraction times pi.

Pi Han Goh - 3 years, 10 months ago

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@Pi Han Goh But for the purpose of this question, we just need to determine a solution so that X X is a positive integer

Noel Lo - 3 years, 10 months ago

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Yes, I know. But how do you know that 30 is the only positive integer solution?

Pi Han Goh - 3 years, 10 months ago

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@Pi Han Goh – Ok, let's start with 1 2 + cos ⁡ ( 2 ( x + 1 6 π ) ) = cos ⁡ ( x + 1 6 π ) 2 cos ⁡ 2 ( x + 1 6 π ) − cos ⁡ ( x + 1 6 π ) − 1 2 = 0 [ 2 cos ⁡ ( x + 1 6 π ) − 1 2 ] 2 = 5 4 cos ⁡ ( x + 1 6 π ) = 1 ± 5 4 = cos ⁡ 1 5 π , cos ⁡ 3 5 π \begin{aligned} \tfrac12 + \cos\big(2(x+\tfrac16\pi)\big) & = \; \cos(x + \tfrac16\pi) \\ 2\cos^2(x + \tfrac16\pi) - \cos(x + \tfrac16\pi) - \tfrac12 & = \; 0 \\ \big[2\cos(x + \tfrac16\pi) - \tfrac12\big]^2 & = \; \tfrac54 \\ \cos(x + \tfrac16\pi) & = \; \frac{1 \pm \sqrt{5}}{4} \; = \; \cos\tfrac15\pi\,,\,\cos\tfrac35\pi \end{aligned} The general solution of cos ⁡ A = cos ⁡ B \cos A = \cos B is A = 2 n π ± B A = 2n\pi \pm B for any integer n n . Thus x + 1 6 π = 2 n π ± 1 5 π , 2 n π ± 3 5 π x = 1 30 π + 2 n π , − 11 30 + 2 n π , 13 30 π + 2 n π , − 23 30 π + 2 n π \begin{aligned} x + \tfrac16\pi & = \; 2n\pi \pm \tfrac15\pi \,,\, 2n\pi \pm \tfrac35\pi \\ x & = \; \tfrac{1}{30}\pi + 2n\pi\,,\,-\tfrac{11}{30} + 2n\pi\,,\,\tfrac{13}{30}\pi + 2n\pi\,,\,-\tfrac{23}{30}\pi + 2n\pi \end{aligned} for any integer n n .

Mark Hennings - 3 years, 10 months ago

Noel, we also want to ensure that the answer is unique. As I mentioned, there are multiple values of (small) x x that would work. You are correct in the claim that there is only 1 unique value of X X that would work. However, that has not been proven (nor stated) in your solution.

Do you understand the gap that you're making? As I mentioned below, this is a very common mistake/misconception when solving trigonometric equations.

Calvin Lin Staff - 3 years, 10 months ago
Chew-Seong Cheong
Aug 12, 2017

1 + sin ⁡ x + cos ⁡ 2 x sin ⁡ 2 x + cos ⁡ x = 3 1 + sin ⁡ x + cos ⁡ 2 x = 3 sin ⁡ 2 x + 3 cos ⁡ x 1 + sin ⁡ x − 3 cos ⁡ x = 3 sin ⁡ 2 x − cos ⁡ 2 x 1 + 2 ( 1 2 sin ⁡ x − 3 2 cos ⁡ x ) = 2 ( 3 2 sin ⁡ 2 x − 1 2 cos ⁡ 2 x ) 1 − 2 cos ⁡ ( x + π 6 ) = − 2 cos ⁡ ( 2 x + π 3 ) Let θ = x + π 6 1 − 2 cos ⁡ θ = − 2 cos ⁡ 2 θ cos ⁡ θ − cos ⁡ 2 θ = 1 2 \begin{aligned} \frac {1+\sin x + \cos 2x}{\sin 2x + \cos x} & = \sqrt 3 \\ 1+\sin x + \cos 2x & = \sqrt 3 \sin 2x + \sqrt 3 \cos x \\ 1 + \sin x - \sqrt 3 \cos x & = \sqrt 3 \sin 2x - \cos 2x \\ 1 + 2 \left(\frac 12 \sin x - \frac {\sqrt 3}2 \cos x \right) & = 2 \left(\frac {\sqrt 3}2 \sin 2x - \frac 12 \cos 2x \right) \\ 1 - 2 \cos \left(x + \frac \pi 6 \right) & = - 2 \cos \left(2x + \frac \pi 3 \right) & \small \color{#3D99F6} \text{Let }\theta = x + \frac \pi 6 \\ 1 - 2\cos \theta & = -2\cos 2\theta \\ \cos \theta - \cos 2 \theta & = \frac 12 \end{aligned}

Note that ∑ k = 0 n − 1 cos ⁡ ( 2 k + 1 2 n + 1 π ) = 1 2 \displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12 (see proof ). A particular case is 2 n + 1 = 5 2n+1=5 , then cos ⁡ π 5 + cos ⁡ 3 π 5 = 1 2 \cos \dfrac \pi 5 + \cos \dfrac {3\pi}5 = \dfrac 12 .

We note that θ = π 5 \theta = \dfrac \pi 5 is a solution, because then cos ⁡ 2 θ = cos ⁡ 2 π 5 = − cos ⁡ 3 π 5 \cos 2\theta = \cos \dfrac {2\pi} 5 = - \cos \dfrac {3\pi}5 . ⟹ cos ⁡ θ − cos ⁡ 2 θ = cos ⁡ π 5 + cos ⁡ 3 π 5 = 1 2 \implies \cos \theta - \cos 2\theta = \cos \dfrac \pi 5 + \cos \dfrac {3\pi}5 = \dfrac 12 . And from θ = x + π 6 = π 5 \theta = x + \dfrac \pi 6 = \dfrac \pi 5 , ⟹ x = π 30 \implies x = \dfrac \pi{30} .

⟹ X = 30 \implies X = \boxed{30} .

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