Difference between revisions of "1991 AIME Problems/Problem 9"
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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 | + | If we square the given <math>\sec x = \frac{22}{7} - \tan x</math>, we find that |
− | + | <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> | ||
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== 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 15:04, 3 August 2008
Problem
Suppose that and that where is in lowest terms. Find
Contents
[hide]Solution
Solution 1
Use the two trigonometric Pythagorean identities and .
If we square the given , we find that
This yields .
Let . Then squaring,
Substituting yields a quadratic equation: . It turns out that only the positive root will work, so the value of and .
Solution 2
Recall that , from which we find that . Adding the equations
together and dividing by 2 gives , and subtracting the equations and dividing by 2 gives . Hence, and . Thus, and . Finally,
so .
Solution 3
(least computation) By the given, and .
Multiplying the two, we have
Subtracting both of the two given equations from this, and simpliyfing with the identity , we get
Solving yields , and
Solution 4
Make the substitution (a substitution commonly used in calculus). , so . Now note the following:
Plugging these into our equality gives:
This simplifies to , and solving for gives , and . Finally, .
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 |