Did Trigonometry just fail?

Algebra Level 3

If a + 1 a = 2 cos 6 a + \dfrac 1a = 2\cos 6^\circ , find a 1000 + 1 a 1000 + 1 a^{ 1000 } +\dfrac 1{a^{1000}}+1 .

0 1 -1 2

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2 solutions

a + 1 a = 2 cos 6 a 2 2 cos 6 a + 1 = 0 a+\dfrac{1}{a} = 2\cos{6^\circ} \quad \Rightarrow a^2 - 2 \cos{6^\circ}a + 1 = 0 a = 2 cos 6 ± 4 cos 2 6 4 2 = cos 6 ± i sin 6 = e ± 6 i \Rightarrow a = \dfrac {2\cos{6^\circ} \pm \sqrt{4\cos^2{6^\circ}-4}}{2} = \cos{6^\circ} \pm i \sin{6^\circ} = e^{\pm6^\circ i}

a 1000 + 1 a 1000 + 1 = e 600 0 i + e 600 0 i + 1 = 2 cos 600 0 + 1 = 2 cos 24 0 + 1 = 2 ( 1 2 ) + 1 = 1 + 1 = 0 \begin{aligned} \Rightarrow a^{1000}+\dfrac{1}{a^{1000}} + 1 & = e^{6000^\circ i} +e^{-6000^\circ i} + 1 = 2\cos {6000^\circ } + 1 \\ & = 2\cos {240^\circ } + 1 = 2 \left( - \frac {1}{2} \right) + 1 \\ & = -1 + 1 = \boxed{0} \end{aligned}

It really a nice question.

ABHIJIT DIXIT - 3 years, 5 months ago

I want a redo lol. I had the same exact solution, but I forgot to add the 1.

James Wilson - 3 years, 5 months ago

I did De Moivre's Theorem

Hans Gabriel Daduya - 3 years, 5 months ago

very nice solution!

I Gede Arya Raditya Parameswara - 3 years, 5 months ago

What about e^-6i?

Rohan Joshi - 4 months, 2 weeks ago

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If a = e 6 i a = e^{-6i} , then a 1000 + 1 a 1000 + 1 = e 6000 i + e 6000 i + 1 a^{1000} + \dfrac 1{a^{1000}} + 1 = e^{-6000i} + e^{6000i} + 1 which is the same expression.

Chew-Seong Cheong - 4 months, 2 weeks ago

Did the same way exactly.It was a cute and a nice question.

rahul saxena - 6 years, 2 months ago
Omkar Kulkarni
Mar 30, 2015

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