Difference between revisions of "2007 AMC 12A Problems/Problem 17"
Armalite46 (talk | contribs) m (→Solution) |
m (fixed link) |
||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | We can make use the of the trigonometric [[Pythagorean identities]]: square both equations and add them up: | + | We can make use the of the trigonometric [[Trigonometric identities#Pythagorean Identities|Pythagorean identities]]: square both equations and add them up: |
<div style="text-align:center;"><math>\sin^2 a + \sin^2 b + 2\sin a \sin b + \cos^2 a + \cos^2 b + 2\cos a \cos b = \frac{5}{3} + 1</math><br /><math>2 + 2\sin a \sin b + 2\cos a \cos b = \frac{8}{3}</math><br /><math>2(\cos a \cos b + \sin a \sin b) = \frac{2}{3}</math></div> | <div style="text-align:center;"><math>\sin^2 a + \sin^2 b + 2\sin a \sin b + \cos^2 a + \cos^2 b + 2\cos a \cos b = \frac{5}{3} + 1</math><br /><math>2 + 2\sin a \sin b + 2\cos a \cos b = \frac{8}{3}</math><br /><math>2(\cos a \cos b + \sin a \sin b) = \frac{2}{3}</math></div> |
Latest revision as of 14:56, 7 August 2017
Problem
Suppose that and
. What is
?
Solution
We can make use the of the trigonometric Pythagorean identities: square both equations and add them up:
![$\sin^2 a + \sin^2 b + 2\sin a \sin b + \cos^2 a + \cos^2 b + 2\cos a \cos b = \frac{5}{3} + 1$](http://latex.artofproblemsolving.com/5/a/6/5a6d5de20ab0bc919b7a7858544f89155541d275.png)
![$2 + 2\sin a \sin b + 2\cos a \cos b = \frac{8}{3}$](http://latex.artofproblemsolving.com/f/2/f/f2f7e9b21268fc2f419dabe37d0440e8e82c6ed3.png)
![$2(\cos a \cos b + \sin a \sin b) = \frac{2}{3}$](http://latex.artofproblemsolving.com/6/f/6/6f689d46f91ed11c75ff9271ff4741e35ba035df.png)
This is just the cosine difference identity, which simplifies to
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.