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2004 AMC 12A Problem 21

Below is the professionally curated solution for Problem 21 of the 2004 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12A solutions, or check the answer key.

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Concepts:geometric sequencetrigonometric identity

Difficulty rating: 1820

21.

If ∑n=0∞cos⁡2nθ=5,\sum_{n=0}^{\infty} \cos^{2n} \theta = 5, what is the value of cos⁡2θ?\cos 2\theta?

15\dfrac{1}{5}

25\dfrac{2}{5}

55\dfrac{\sqrt{5}}{5}

35\dfrac{3}{5}

45\dfrac{4}{5}

Solution:

The series is geometric with first term 11 and ratio cos⁡2θ,\cos^2 \theta, so its sum is 11−cos⁡2θ=1sin⁡2θ=5.\dfrac{1}{1 - \cos^2 \theta} = \dfrac{1}{\sin^2 \theta} = 5.

Thus sin⁡2θ=15,\sin^2 \theta = \tfrac15, and cos⁡2θ=1−2sin⁡2θ=1−25=35. \cos 2\theta = 1 - 2\sin^2 \theta = 1 - \tfrac25 = \tfrac35.

Thus, the correct answer is D.

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