Difference between revisions of "1991 AIME Problems/Problem 9"
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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 19:53, 24 March 2009
Problem
Suppose that and that
where
is in lowest terms. Find
Contents
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,
.
Solution 5
We are given that
, or equivalently,
. Note that what we want is just
.
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 |