2010 AMC 12B Problems/Problem 20: Difference between revisions
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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 | ||
< | <cmath>a_1= \sin x</cmath> | ||
< | <cmath>a_2= \cos x</cmath> | ||
< | <cmath>a_3= \frac{\cos^2x}{\sin x}</cmath> | ||
< | <cmath>a_4=\frac{\cos^3x}{\sin^2x}=1</cmath> | ||
< | <cmath>a_5=\frac{\cos x}{\sin x}</cmath> | ||
< | <cmath>a_6=\frac{\cos^2x}{\sin^2x}</cmath> | ||
< | <cmath>a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}</cmath> | ||
< | <cmath>a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}</cmath> | ||
< | <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>. | ||
== 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 
has 
, 
, and 
for some real number 
. For what value of 
does 
?
Solution
By defintion of a geometric sequence, we have 
. Since 
, we can rewrite this as 
.
The common ratio of the sequence is 
, so we can write
![\[a_1= \sin x\]](http://latex.artofproblemsolving.com/1/a/a/1aabb1dd85adb650024782c4a00d25af69ce9d08.png)
![\[a_2= \cos x\]](http://latex.artofproblemsolving.com/4/3/c/43ca1cff036216ab2bfd7f46fbdb3e5c9fd953cd.png)
![\[a_3= \frac{\cos^2x}{\sin x}\]](http://latex.artofproblemsolving.com/b/2/3/b2393d9ecc7f124d48e1f02f9b4e73260bff9d5a.png)
![\[a_4=\frac{\cos^3x}{\sin^2x}=1\]](http://latex.artofproblemsolving.com/1/9/2/192747714a5472598036838a6af4ca461a7fb764.png)
![\[a_5=\frac{\cos x}{\sin x}\]](http://latex.artofproblemsolving.com/2/0/3/203672b722b8e4a64a0f72c6e2271938abe92355.png)
![\[a_6=\frac{\cos^2x}{\sin^2x}\]](http://latex.artofproblemsolving.com/6/1/0/6105eae8dce61679696242306072701b1a158834.png)
![\[a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}\]](http://latex.artofproblemsolving.com/9/a/f/9af8c8f10513413e48d001ab1bc43bac716fb4f9.png)
![\[a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}\]](http://latex.artofproblemsolving.com/6/7/f/67fb2736c525445714ed24588d8b56676298dc00.png)
![\[a_9=\frac{\cos x}{\sin x}\]](http://latex.artofproblemsolving.com/4/7/7/477aecd5abf5f1282dd7ece181bf25de36a05cb9.png)
We can conclude that the sequence from 
to 
repeats.
Since 
, we have 
, which is making our answer 
.
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 | |
