Difference between revisions of "2007 AMC 12A Problems/Problem 17"
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== Solution == | == Solution == | ||
| − | We can make use the of the [[Pythagorean identities]]: square both equations and add them up: | + | We can make use the of the trigonometric [[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> | ||
Revision as of 19:43, 27 October 2013
Problem
Suppose that 
and 
. What is 
?
Solution
We can make use the of the trigonometric Pythagorean identities: square both equations and add them up:
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. 
