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Difference between revisions of "1991 AIME Problems/Problem 9"

(fmtt)
(solution (2) by chess64, solution (3) by dgreenb801)
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Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.  
 
Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.  
  
−
If we square <math>\sec x = \frac{22}{7} - \tan x</math>, we find that <math>\sec^2 x = \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x</math>, so <math>1 = \left(\frac{22}7\right)^2 - \frac{44}7 \tan x</math>. Solving shows that <math>\tan x = \frac{435}{308}</math>.
+
If we square the given <math>\sec x = \frac{22}{7} - \tan x</math>, we find that  
  
−
Call <math>y = \frac mn</math>. Rewrite the second equation in a similar fashion: <math>1 = y^2 - 2y\cot x</math>. Substitute in <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> to get a [[quadratic equation|quadratic]]: <math>0 = y^2 - \frac{616}{435}y - 1</math>. This factors as <math>(15y - 29)(29y + 15) = 0</math>. It turns out that only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = \boxed{044}</math>.
+
<cmath>\begin{align*}
 +
\sec^2 x &= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\
 +
1 &= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}</cmath>
  
−
=== Solution 2===
+
This yields <math>\tan x = \frac{435}{308}</math>.
 +
 
 +
Let <math>y = \frac mn</math>. Then squaring,
 +
 
 +
<cmath>\csc^2 x = (y - \cot^2 x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.</cmath>
 +
 
 +
Substituting <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> yields a [[quadratic equation]]: <math>0 = 435y^2 - \frac{616}{435}y - 435 = 15y - 29)(29y + 15)</math>. It turns out that only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = \boxed{044}</math>.
 +
 
 +
=== Solution 2 ===
 +
Recall that <math>\sec^2 x - \tan^2 x = 1</math>, from which we find that <math>\sec x - \tan x = 7/22</math>. Adding the equations
 +
 
 +
<cmath>\begin{eqnarray*} \sec x + \tan x & = & 22/7 \\
 +
\sec x - \tan x & = & 7/22\end{eqnarray*}</cmath>
 +
 
 +
together and dividing by 2 gives <math>\sec x = 533/308</math>, and subtracting the equations and dividing by 2 gives <math>\tan x = 435/308</math>. Hence, <math>\cos x = 308/533</math> and <math>\sin x = \tan x \cos x = (435/308)(308/533) = 435/533</math>. Thus, <math>\csc x = 533/435</math> and <math>\cot x = 308/435</math>. Finally,
 +
 
 +
<cmath>\csc x + \cot x = \frac {841}{435} = \frac {29}{15},</cmath>
 +
 
 +
so <math>m + n = 044</math>.
 +
 
 +
=== Solution 3 ===
 +
(least computation) By the given,
 +
<math>\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}</math> and
 +
<math>\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k</math>.
 +
 
 +
Multiplying the two, we have
 +
 
 +
<cmath>\frac {1}{\sin x \cos x} + \frac {1}{\sin x} + \frac {1}{\cos x} + 1 = \frac {22}{7}k</cmath>
 +
 
 +
Subtracting both of the two given equations from this, and simpliyfing with the identity <math>\frac {\sin x}{\cos x} + \frac {\cos x}{\sin x} = \frac{\sin ^2 x + \cos ^2 x}{\sin x \cos x} = \frac {1}{\sin x \cos x}</math>, we get
 +
 
 +
<cmath>1 = \frac {22}{7}k - \frac {22}{7} - k.</cmath>
 +
 
 +
Solving yields <math>k = \frac {29}{15}</math>, and <math>m+n = 044</math>
 +
 
 +
=== Solution 4 ===
 
Make the substitution <math>u = \tan \frac x2</math> (a substitution commonly used in calculus). <math>\tan \frac x2 = \frac{\sin x}{1+\cos x}</math>, so <math>\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn</math>. <math>\sec x + \tan x = \frac{1 + \sin x}{\cos x}.</math> Now note the following:
 
Make the substitution <math>u = \tan \frac x2</math> (a substitution commonly used in calculus). <math>\tan \frac x2 = \frac{\sin x}{1+\cos x}</math>, so <math>\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn</math>. <math>\sec x + \tan x = \frac{1 + \sin x}{\cos x}.</math> Now note the following:
 +
 
<cmath>\begin{align*}\sin x &= \frac{2u}{1+u^2}\\
 
<cmath>\begin{align*}\sin x &= \frac{2u}{1+u^2}\\
 
\cos x &= \frac{1-u^2}{1+u^2}\end{align*}</cmath>
 
\cos x &= \frac{1-u^2}{1+u^2}\end{align*}</cmath>
 +
 
Plugging these into our equality gives:
 
Plugging these into our equality gives:
 +
 
<cmath>\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7</cmath>
 
<cmath>\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7</cmath>
  
Line 21: Line 61:
  
 
== See also ==
 
== See also ==
−
{{AIME box|year=1991|num-b=8|num-a=10}}
+
{{AIME box|year=1991|num-b=8|num-a=10|t=8322}}
  
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]

Revision as of 16:04, 3 August 2008

Problem

Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in lowest terms. Find $m+n^{}_{}.$

Solution

Solution 1

Use the two trigonometric Pythagorean identities $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$.

If we square the given $\sec x = \frac{22}{7} - \tan x$, we find that

\begin{align*} \sec^2 x &= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\ 1 &= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}

This yields $\tan x = \frac{435}{308}$.

Let $y = \frac mn$. Then squaring,

\[\csc^2 x = (y - \cot^2 x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.\]

Substituting $\cot x = \frac{1}{\tan x} = \frac{308}{435}$ yields a quadratic equation: $0 = 435y^2 - \frac{616}{435}y - 435 = 15y - 29)(29y + 15)$. It turns out that only the positive root will work, so the value of $y = \frac{29}{15}$ and $m + n = \boxed{044}$.

Solution 2

Recall that $\sec^2 x - \tan^2 x = 1$, from which we find that $\sec x - \tan x = 7/22$. Adding the equations

\begin{eqnarray*} \sec x + \tan x & = & 22/7 \\ \sec x - \tan x & = & 7/22\end{eqnarray*}

together and dividing by 2 gives $\sec x = 533/308$, and subtracting the equations and dividing by 2 gives $\tan x = 435/308$. Hence, $\cos x = 308/533$ and $\sin x = \tan x \cos x = (435/308)(308/533) = 435/533$. Thus, $\csc x = 533/435$ and $\cot x = 308/435$. Finally,

\[\csc x + \cot x = \frac {841}{435} = \frac {29}{15},\]

so $m + n = 044$.

Solution 3

(least computation) By the given, $\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}$ and $\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k$.

Multiplying the two, we have

\[\frac {1}{\sin x \cos x} + \frac {1}{\sin x} + \frac {1}{\cos x} + 1 = \frac {22}{7}k\]

Subtracting both of the two given equations from this, and simpliyfing with the identity $\frac {\sin x}{\cos x} + \frac {\cos x}{\sin x} = \frac{\sin ^2 x + \cos ^2 x}{\sin x \cos x} = \frac {1}{\sin x \cos x}$, we get

\[1 = \frac {22}{7}k - \frac {22}{7} - k.\]

Solving yields $k = \frac {29}{15}$, and $m+n = 044$

Solution 4

Make the substitution $u = \tan \frac x2$ (a substitution commonly used in calculus). $\tan \frac x2 = \frac{\sin x}{1+\cos x}$, so $\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn$. $\sec x + \tan x = \frac{1 + \sin x}{\cos x}.$ Now note the following:

\begin{align*}\sin x &= \frac{2u}{1+u^2}\\ \cos x &= \frac{1-u^2}{1+u^2}\end{align*}

Plugging these into our equality gives:

\[\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7\]

This simplifies to $\frac{1+u}{1-u} = \frac{22}7$, and solving for $u$ gives $u = \frac{15}{29}$, and $\frac mn = \frac{29}{15}$. Finally, $m+n = 044$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions
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