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2010 AMC 12B Problems/Problem 20

Problem

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 $a_n=1+\cos x$?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

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 $8 \Rightarrow \boxed{E}$.

Solution 2

Notice that the common ratio is $r=\frac{\cos(x)}{\sin(x)}$; multiplying it to $\tan(x)=\frac{\sin(x)}{\cos(x)}$ gives $a_4=1$. Then, working backwards we have $a_3=\frac{1}{r}$, $a_2=\frac{1}{r^2}$ and $a_1=\frac{1}{r^3}$. Now notice that since $a_1=\sin(x)$ and $a_2=\cos(x)$, we need $a_1^2+a_2^2=1$, so $\frac{1}{r^6}+\frac{1}{r^4}=\frac{r^2+1}{r^6}=1\implies r^2+1=r^6$. Dividing both sides by $r^2$ gives $1+\frac{1}{r^2}=r^4$, which the left side is equal to $1+\cos(x)$; we see as well that the right hand side is equal to $a_8$ given $a_4=1$, so the answer is $\boxed{E}$. - mathleticguyyy

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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