Difference between revisions of "2007 AMC 12A Problems/Problem 17"

(See also)
m (fixed link)
 
(One intermediate revision by one other user not shown)
Line 5: Line 5:
  
 
== 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 [[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 13:56, 7 August 2017

Problem

Suppose that $\sin a + \sin b = \sqrt{\frac{5}{3}}$ and $\cos a + \cos b = 1$. What is $\cos (a - b)$?

$\mathrm{(A)}\ \sqrt{\frac{5}{3}} - 1\qquad \mathrm{(B)}\ \frac 13\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 23\qquad \mathrm{(E)}\ 1$

Solution

We can make use the of the trigonometric Pythagorean identities: square both equations and add them up:

$\sin^2 a + \sin^2 b + 2\sin a \sin b + \cos^2 a + \cos^2 b + 2\cos a \cos b = \frac{5}{3} + 1$
$2 + 2\sin a \sin b + 2\cos a \cos b = \frac{8}{3}$
$2(\cos a \cos b + \sin a \sin b) = \frac{2}{3}$

This is just the cosine difference identity, which simplifies to $\cos (a - b) = \frac{1}{3} \Longrightarrow \mathrm{(B)}$

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png