Difference between revisions of "1995 AIME Problems/Problem 7"
(→Solution 2) |
(→Solution 2) |
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We want to find <math>1+\sin x \cos x-\sin x-\cos x</math> | We want to find <math>1+\sin x \cos x-\sin x-\cos x</math> | ||
− | If we find \sin x+\cos x, we will be done with the problem. | + | If we find <math>\sin x+\cos x</math>, we will be done with the problem. |
Let <math>y = \sin x+\cos x</math> | Let <math>y = \sin x+\cos x</math> |
Revision as of 12:53, 16 June 2018
Contents
Problem
Given that and
where and
are positive integers with
and
relatively prime, find
Solution
From the givens,
, and adding
to both sides gives
. Completing the square on the left in the variable
gives
. Since
, we have
. Subtracting twice this from our original equation gives
, so the answer is
.
Solution 2
Let . Multiplying
with the given equation,
, and
. Simplifying and rearranging the given equation,
. Notice that
, and substituting,
. Rearranging and squaring,
, so
, and
, but clearly,
. Therefore,
, and the answer is
.
==SOLUTION 3 == (-synergy)
We have
We want to find
If we find , we will be done with the problem.
Let
Squaring, we have
From this we have and
Substituting this into the first equation we have ,
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.