2004 AMC 12A 第 21 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

21.

如果 那么 cos2θ\cos 2\theta 的值是多少? n=0cos2nθ=5,\sum_{n=0}^{\infty} \cos^{2n} \theta = 5,

If n=0cos2nθ=5,\sum_{n=0}^{\infty} \cos^{2n} \theta = 5, what is the value of cos2θ?\cos 2\theta?

15\dfrac{1}{5}

25\dfrac{2}{5}

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

35\dfrac{3}{5}

45\dfrac{4}{5}

答案:D
知识点:等比数列三角恒等式
难度评级:1820
解答:

该级数首项为 11,公比为 cos2θ\cos^2 \theta,所以和为 11cos2θ=1sin2θ=5\dfrac{1}{1 - \cos^2 \theta} = \dfrac{1}{\sin^2 \theta} = 5

因此 sin2θ=15\sin^2 \theta = \tfrac15,并且 cos2θ=12sin2θ=125=35. \begin{aligned} \cos 2\theta &= 1 - 2\sin^2 \theta \\ &= 1 - \tfrac25 = \tfrac35. \end{aligned}

所以正确答案是 D

The series is geometric with first term 11 and ratio cos2θ,\cos^2 \theta, so its sum is 11cos2θ=1sin2θ=5.\dfrac{1}{1 - \cos^2 \theta} = \dfrac{1}{\sin^2 \theta} = 5.

Thus sin2θ=15,\sin^2 \theta = \tfrac15, and cos2θ=12sin2θ=125=35. \begin{aligned} \cos 2\theta &= 1 - 2\sin^2 \theta \\ &= 1 - \tfrac25 = \tfrac35. \end{aligned}

Thus, the correct answer is D.

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