AMC12 2004 A

AMC12 2004 A · Q21

AMC12 2004 A · Q21. It mainly tests Exponents & radicals, Functions basics.

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

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

(A) $\dfrac{1}{5}$ $\dfrac{1}{5}$

(B) $\dfrac{2}{5}$ $\dfrac{2}{5}$

(D) $\dfrac{3}{5}$ $\dfrac{3}{5}$

(E) $\dfrac{4}{5}$ $\dfrac{4}{5}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

This is an infinite geometric series, which sums to $\frac{\cos^0 \theta}{1 - \cos^2 \theta} = 5 \Longrightarrow 1 = 5 - 5\cos^2 \theta \Longrightarrow \cos^2 \theta = \frac{4}{5}$. Using the formula $\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}$. We can more directly solve this with superficially less work. Again, applying the formula for an infinite geometric series, \[\sum_{i=0}^{\infty}\cos^{2i}\theta=\dfrac1{1-\cos^2\theta}=\dfrac1{\sin^2\theta}=5.\] Thus, $\sin^2\theta=\dfrac15$, so $\cos(2\theta)=1-2\sin^2\theta=1-\dfrac25=\dfrac35.$

这是一个无穷等比级数，其和为 $\frac{\cos^0 \theta}{1 - \cos^2 \theta} = 5 \Longrightarrow 1 = 5 - 5\cos^2 \theta \Longrightarrow \cos^2 \theta = \frac{4}{5}$。利用公式 $\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}$。我们也可以用表面上更少的步骤更直接地解。仍然应用无穷等比级数求和公式， \[\sum_{i=0}^{\infty}\cos^{2i}\theta=\dfrac1{1-\cos^2\theta}=\dfrac1{\sin^2\theta}=5.\] 因此，$\sin^2\theta=\dfrac15$，所以 $\cos(2\theta)=1-2\sin^2\theta=1-\dfrac25=\dfrac35.$ 证毕。

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