Difference between revisions of "2002 AMC 12P Problems/Problem 25"

Latest revision as of 20:02, 3 October 2024

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 $\boxed{(C)}$ .

~alexanderruan

2002 AMC 12P (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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@@ Line 15: / Line 15: @@
 == Solution ==
-Suppose we substitute <math>\frac{a+b}{2} = x</math> and <math>\frac{a-b}{2} = y</math>. Sum to product gives us
+Sum to product gives us
-<cmath>2\sin{x}\cos{y} = \frac{\sqrt{2}}{2}</cmath>
+<cmath>2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{2}}{2}</cmath>
-<cmath>2\cos{x}\cos{y} = \frac{\sqrt{6}}{2}.</cmath>
+<cmath>2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2}) = \frac{\sqrt{6}}{2}</cmath>
-Dividing these equations tells us that <math>\tan{x} = \frac{1}{\sqrt{3}}</math>, so <math>x = \frac{\pi}{6} + \pi n</math> for an integer <math>n</math>. Note that <math>a+b = 2x</math>, so <math>\sin{a+b} = \sin{2x} = \sin{\pi}{3} + 2\pi n = \frac{\sqrt{3}}{2}</math>, so our answer is <math>\boxed{B}</math>.
+Dividing these equations tells us that <math>\tan(\frac{a+b}{2}) = \frac{1}{\sqrt{3}}</math>, so <math>\frac{a+b}{2} = \frac{\pi}{6} + \pi n</math> for an integer <math>n</math>, so <math>\sin(a+b) = \sin (\frac{\pi}{3} + 2\pi n) = \frac{\sqrt{3}}{2}</math>. The answer is <math>\boxed{(C)}</math>.
-== Solution 2 (doesn't work but gives the right answer) ==
+~alexanderruan
-Given <math>\begin{cases}\sin{a} + \sin{b} = \frac{\sqrt{2}}{2} \dots \textcircled{1}\\
-\cos {a} + \cos {b} = \frac{\sqrt{6}}{2} \dots \textcircled{2}\end{cases} </math>
-We multiply both sides of the syetem, <math>\textcircled{1} \times \textcircled{2}</math>, then  we get <math> (\sin{a}\cos{a}  + \sin{b} \cos{b}  )+( \sin{a}\cos{b}  + \sin{b} \cos{a} )= \frac{\sqrt{3}}{2}</math>. i.e.
-<math>(\sin{a}\cos{a}  + \sin{b} \cos{b}  )+\sin{(a+b)}= \frac{\sqrt{3}}{2}</math>.
-We must get the sum of the first part of the equation, then we calculate <math>\textcircled{1}^2+\textcircled{2}^2 </math>, we will
-get <math>\sin{a}\cos{a}  + \sin{b} \cos{b} = 0 </math> as <math> \sin^{2}{a}+\cos^{2}{a} = 1</math> and <math> \sin^{2}{b}+\cos^{2}{b} = 1</math>.
-So <math>\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\textbf{(C) }\frac{\sqrt{3}}{2}}</math>
-Comment: This problem is pretty much identical to [[2007 AMC 12A Problems/Problem 17|2007 AMC 12A Problem 17]] except with different numbers.
-Note: This solution is wrong since equation 1 square plus equation 2 squared gives sin a sin b and cos a cos b.
 == See also ==
 {{AMC12 box|year=2002|ab=P|num-b=24|after=Last question}}
 {{MAA Notice}}

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Difference between revisions of "2002 AMC 12P Problems/Problem 25"

Latest revision as of 20:02, 3 October 2024

Problem

Solution

See also