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Geometry Level 4

If sin x + sin y + sin z = cos x + cos y + cos z = 0 \sin x+\sin y+ \sin z =\cos x+\cos y+\cos z=0 , then find ( cos x ) 2 + ( cos y ) 2 + ( cos z ) 2 . (\cos x)^2 +(\cos y)^2 +(\cos z )^2.

1.2 0 1.5 1.4

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1 solution

sin x + sin y + sin z = cos x + cos y + cos z = 0 cos x + cos y + cos z + i ( sin x + sin y + sin z ) = 0 e i x + e i y + e i z = 0 An trivial solution is the third roots of unity ω 2 + ω + 1 = 0 e i 4 π 3 + e i 2 π 3 + e i 0 = 0 x , y , z = 0 , 2 π 3 , 4 π 3 cos 2 x + cos 2 y + cos 2 z = cos 2 0 + cos 2 2 π 3 + cos 2 4 π 3 = 1 2 + ( 1 2 ) 2 + ( 1 2 ) 2 = 1.5 \begin{aligned} \sin x + \sin y + \sin z = \cos x + \cos y + \cos z & = 0 \\ \Rightarrow \cos x + \cos y + \cos z + i(\sin x + \sin y + \sin z) & = 0 \\ e^{ix} + e^{iy} + e^{iz} & = 0\\ \color{#3D99F6}{\text{An trivial solution is the third roots of unity}} \quad \Rightarrow \omega^2 + \omega + 1 & = 0 \\ e^{i\frac{4\pi}{3}} + e^{i\frac{2\pi}{3}} + e^{i0} & = 0 \\ \Rightarrow x, y, z & = 0, \frac{2\pi}{3}, \frac{4 \pi} {3} \\ \Rightarrow \cos^2 x + \cos^2 y + \cos^2 z & = \cos^2 0 + \cos^2 \frac{2\pi}{3} + \cos^2 \frac{4\pi}{3} \\ & = 1^2 + \left(-\frac{1}{2} \right)^2 + \left(-\frac{1}{2} \right)^2 = \boxed{1.5} \end{aligned}

By your solution sin(x)+sin(y)+sin(z) is not equal to zero

Niketan Nath - 5 years, 8 months ago

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You are right. It should be 0 , 2 π 3 , 4 π 3 0, \frac{2 \pi}{3}, \frac{4 \pi}{3} .

Chew-Seong Cheong - 5 years, 8 months ago

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Yes, now it is correct

Niketan Nath - 5 years, 8 months ago

hello sir this is not actually a solution, it's a trial by error method

Benjamin ononogbu - 5 years, 5 months ago

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Yes, let's wait for a better solution.

Chew-Seong Cheong - 5 years, 5 months ago

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