2010 AMC 12B Problems/Problem 20: Difference between revisions

Revision as of 13:50, 3 April 2016

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 $8 \Rightarrow \boxed{E}$ .

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

@@ Line 20: / Line 20: @@
-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 ==
 {{AMC12 box|year=2010|num-b=19|num-a=21|ab=B}}
 {{MAA Notice}}

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2010 AMC 12B Problems/Problem 20: Difference between revisions

Revision as of 13:50, 3 April 2016

Problem

Solution

See also