Why does one trigonometry problem matter

Geometry Level 2

Given that U n = 2 cos ⁡ n θ U_n=2\cos n\theta

Find the value of U 1 U n − U n − 1 U_1U_n-U_{n-1}

U n U_n U n + 2 U_{n+2} U n + 1 U_{n+1} U − n U_{-n}

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Vaibhav Prasad by Vaibhav Prasad , 21, India

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1 solution

Chew-Seong Cheong
Jul 18, 2015

U 1 U n − U n − 1 = 2 cos ⁡ θ ( 2 cos ⁡ n θ ) − 2 cos ⁡ ( [ n − 1 ] θ ) = 4 cos ⁡ θ cos ⁡ n θ − 2 ( cos ⁡ n θ cos ⁡ θ + sin ⁡ n θ sin ⁡ θ ) = 2 cos ⁡ n θ cos ⁡ θ − 2 sin ⁡ n θ sin ⁡ θ = 2 cos ⁡ ( [ n + 1 ] θ ) = U n + 1 \begin{aligned} U_1U_n - U_{n-1} & = 2\cos{\theta} (2 \cos{n\theta} ) - 2 \cos{([n-1]\theta)} \\ & = 4\cos{\theta} \cos{n\theta} - 2( \cos{n\theta} \cos{\theta} + \sin{n\theta} \sin{\theta}) \\ & = 2\cos{n\theta} \cos{\theta} - 2\sin{n\theta} \sin{\theta} \\ & = 2\cos{([n+1]\theta)} \\ & = \boxed{U_{n+1}} \end{aligned}

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Moderator note:

Bonus question : Prove that U 2 n = 2 U n 2 − 1 U_{2n} = 2U_n^2 - 1 and U 3 n = 4 U n 3 − 3 U n U_{3n} = 4U_n ^3 - 3U_n . Does these formulas look familiar?

In response to challenger master note: These are the angle half formulas i.e. cos2A and cos3A

Abhay Tiwari - 5 years, 10 months ago

I don't get the proofs.

U 2 n = 2 cos ⁡ 2 n θ = 2 ( 2 cos ⁡ 2 n θ − 1 ) = U n 2 − 2 = U n 2 − U 0 U 3 n = 2 cos ⁡ 3 n θ = 2 ( 4 cos ⁡ 3 n θ − 3 cos ⁡ n θ ) = U n 3 − 3 U n \begin{aligned} U_{2n} & = 2\cos{2n\theta} = 2(2\cos^2{n\theta} -1) = U_n^2 - 2 = U_n^2 -U_0 \\ U_{3n} & = 2\cos{3n\theta} = 2(4\cos^3{n\theta}-3\cos{n\theta}) = U_n^3 - 3U_n \end{aligned}

Chew-Seong Cheong - 5 years, 10 months ago

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