AMC12 2010 B

AMC12 2010 B · Q20

AMC12 2010 B · Q20. It mainly tests Sequences & recursion (algebra), Trigonometry (basic).

A geometric sequence $(a_n)$ has $a_1=\sin x$, $a_2=\cos x$, and $a_3= \tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\cos x$?

一个等比数列 $(a_n)$ 满足 $a_1=\sin x$，$a_2=\cos x$，$a_3= \tan x$，其中 $x$ 为某个实数。问当 $n$ 为多少时有 $a_n=1+\cos x$？

(A) 4 4

(B) 5 5

(D) 7 7

(E) 8 8

Answer

Correct choice: (E)

正确答案：（E）

Solution

By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$. Since $\tan x=\frac{\sin x}{\cos x}$, we can rewrite this as $\cos^3x=\sin^2x$. The common ratio of the sequence is $\frac{\cos x}{\sin x}$, so we can write \[a_1= \sin x\] \[a_2= \cos x\] \[a_3= \frac{\cos^2x}{\sin x}\] \[a_4=\frac{\cos^3x}{\sin^2x}=1\] \[a_5=\frac{\cos x}{\sin x}\] \[a_6=\frac{\cos^2x}{\sin^2x}\] \[a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}\] \[a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}\] Since $\cos^3x=\sin^2x=1-\cos^2x$, we have $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$, which is $a_8$ , making our answer $8 \Rightarrow \boxed{E}$.

由等比数列的定义，有 $\cos^2x=\sin x \tan x$。由于 $\tan x=\frac{\sin x}{\cos x}$，可改写为 $\cos^3x=\sin^2x$。该数列的公比为 $\frac{\cos x}{\sin x}$，因此可写出 \[a_1= \sin x\] \[a_2= \cos x\] \[a_3= \frac{\cos^2x}{\sin x}\] \[a_4=\frac{\cos^3x}{\sin^2x}=1\] \[a_5=\frac{\cos x}{\sin x}\] \[a_6=\frac{\cos^2x}{\sin^2x}\] \[a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}\] \[a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}\] 由于 $\cos^3x=\sin^2x=1-\cos^2x$，有 $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$，这正是 $a_8$，因此答案为 $8 \Rightarrow \boxed{E}$。

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