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

(solution (2) by chess64, solution (3) by dgreenb801)
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This simplifies to <math>\frac{1+u}{1-u} = \frac{22}7</math>, and solving for <math>u</math> gives <math>u = \frac{15}{29}</math>, and <math>\frac mn = \frac{29}{15}</math>. Finally, <math>m+n = 044</math>.
 
This simplifies to <math>\frac{1+u}{1-u} = \frac{22}7</math>, and solving for <math>u</math> gives <math>u = \frac{15}{29}</math>, and <math>\frac mn = \frac{29}{15}</math>. Finally, <math>m+n = 044</math>.
 +
 +
=== Solution 5 ===
 +
We are given that <math>\frac{1+\sin x}{\cos x}=\frac{22}7\implies\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}</math>
 +
<math>=\frac{\cos x}{1-\sin x}</math>, or equivalently, <math>\cos x=\frac{7+7\sin x}{22}=\frac{22-22\sin x}7\implies\sin x=\frac{22^2-7^2}{22^2+7^2}</math>
 +
<math>\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}</math>. Note that what we want is just <math>\frac{1+\cos x}{\sin x}=\frac{1+\frac{2\cdot22\cdot7}{22^2+7^2}}{\frac{22^2-7^2}{22^2+7^2}}=\frac{22^2+7^2+2\cdot22\cdot7}{22^2-7^2}=\frac{(22+7)^2}{(22-7)(22+7)}=\frac{22+7}{22-7}</math>
 +
<math>=\frac{29}{15}\implies m+n=29+15=\boxed{044}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:53, 24 March 2009

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, $m+n = 044$.

Solution 5

We are given that $\frac{1+\sin x}{\cos x}=\frac{22}7\implies\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}$ $=\frac{\cos x}{1-\sin x}$, or equivalently, $\cos x=\frac{7+7\sin x}{22}=\frac{22-22\sin x}7\implies\sin x=\frac{22^2-7^2}{22^2+7^2}$ $\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}$. Note that what we want is just $\frac{1+\cos x}{\sin x}=\frac{1+\frac{2\cdot22\cdot7}{22^2+7^2}}{\frac{22^2-7^2}{22^2+7^2}}=\frac{22^2+7^2+2\cdot22\cdot7}{22^2-7^2}=\frac{(22+7)^2}{(22-7)(22+7)}=\frac{22+7}{22-7}$ $=\frac{29}{15}\implies m+n=29+15=\boxed{044}$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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