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Difference between revisions of "2010 AMC 12B Problems/Problem 20"

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The common ratio of the sequence is <math>\frac{\cos x}{\sin x}</math>, so we can write
 
The common ratio of the sequence is <math>\frac{\cos x}{\sin x}</math>, so we can write
  
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<math>a_1= \sin x</math>
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<cmath>a_1= \sin x</cmath>
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<math>a_2= \cos x</math>
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<cmath>a_2= \cos x</cmath>
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<math>a_3= \frac{\cos^2x}{\sin x}</math>
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<cmath>a_3= \frac{\cos^2x}{\sin x}</cmath>
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<math>a_4=\frac{\cos^3x}{\sin^2x}=1</math>
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<cmath>a_4=\frac{\cos^3x}{\sin^2x}=1</cmath>
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<math>a_5=\frac{\cos x}{\sin x}</math>
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<cmath>a_5=\frac{\cos x}{\sin x}</cmath>
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<math>a_6=\frac{\cos^2x}{\sin^2x}</math>
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<cmath>a_6=\frac{\cos^2x}{\sin^2x}</cmath>
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<math>a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}</math>
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<cmath>a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}</cmath>
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<math>a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}</math>
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<cmath>a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}</cmath>
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<math>a_9=\frac{\cos x}{\sin x}</math>
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<cmath>a_9=\frac{\cos x}{\sin x}</cmath>
  
  
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Since <math>\cos^3x=\sin^2x=1-\cos^2x</math>, we have <math>\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}</math>, which is <math>a_8</math> making our answer <math>8 \Rightarrow \boxed{E}</math>.
 
Since <math>\cos^3x=\sin^2x=1-\cos^2x</math>, we have <math>\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}</math>, which is <math>a_8</math> making our answer <math>8 \Rightarrow \boxed{E}</math>.
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--Please fix formatting--
 
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=19|num-a=21|ab=B}}
 
{{AMC12 box|year=2010|num-b=19|num-a=21|ab=B}}

Revision as of 19:22, 18 August 2011

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 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

\[a_1= \sin x\] \[a_2= \cos x\] \[a_3= \frac{\cos^2x}{\sin x}\] \[a_4=\frac{\cos^3x}{\sin^2x}=1\] \[a_5=\frac{\cos x}{\sin x}\] \[a_6=\frac{\cos^2x}{\sin^2x}\] \[a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}\] \[a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}\] \[a_9=\frac{\cos x}{\sin x}\]


We can conclude that the sequence from $a_4$ to $a_8$ repeats.

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 $8 \Rightarrow \boxed{E}$.

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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