2002 AMC 12P Problems/Problem 25

Problem

Let $a$ and $b$ be real numbers such that $\sin{a} + \sin{b} = \frac{\sqrt{2}}{2}$ and $\cos {a} + \cos {b} = \frac{\sqrt{6}}{2}.$ Find $\sin{(a+b)}.$

$\text{(A) }\frac{1}{2} \qquad \text{(B) }\frac{\sqrt{2}}{2} \qquad \text{(C) }\frac{\sqrt{3}}{2} \qquad \text{(D) }\frac{\sqrt{6}}{2} \qquad \text{(E) }1$

Solution

Sum to product gives us

$2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{2}}{2}$

$2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{6}}{2}$

Dividing these equations tells us that $\tan(\frac{a+b}{2}) = \frac{1}{\sqrt{3}}$ , so $\frac{a+b}{2} = \frac{\pi}{6} + \pi n$ for an integer $n$ , so $\sin(a+b) = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}$ . The answer is $\boxed{\text{(C)}}$ .

~alexanderruan

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last question
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.

Page

Toolbox

Search

2002 AMC 12P Problems/Problem 25

Problem

Solution

See also