Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1972 AHSME Problems/Problem 20"

(Created page with "We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that sour equation is now: <cmath>\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}</cmath> Multiplying through an...")
 
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
−
We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that sour equation is now: <cmath>\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}</cmath> Multiplying through and rearranging gives us the equation: <cmath>\cos x = \frac{a^2-b^2}{2ab} * \sin x</cmath> We now apply the Pythagorean identity <math>\sin ^2 x + \cos ^2 x =1</math>, using our substitution: <cmath>\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1</cmath> We can isolate <math>\sin x</math> without worrying about division by <math>0</math> since <math>a \neq b \neq 0</math> our final answer is <math>(E) \frac{2ab}{a^2+b^2}</math>
+
== Problem 20 ==
 +
 +
If <math>\tan x=\dfrac{2ab}{a^2-b^2}</math> where <math>a>b>0</math> and <math>0^\circ <x<90^\circ</math>, then <math>\sin x</math> is equal to
 +
 
 +
<math>\textbf{(A) }\frac{a}{b}\qquad
 +
\textbf{(B) }\frac{b}{a}\qquad
 +
\textbf{(C) }\frac{\sqrt{a^2-b^2}}{2a}\qquad
 +
\textbf{(D) }\frac{\sqrt{a^2-b^2}}{2ab}\qquad
 +
\textbf{(E) }\frac{2ab}{a^2+b^2} </math>
 +
 
 +
[[1972 AHSME Problems/Problem 20|Solution]]
 +
 
 +
==Solution==
 +
 
 +
We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that our equation is now: <cmath>\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}</cmath> Multiplying through and rearranging gives us the equation: <cmath>\cos x = \frac{a^2-b^2}{2ab} * \sin x</cmath> We now apply the Pythagorean identity <math>\sin ^2 x + \cos ^2 x =1</math>, using our substitution: <cmath>\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1</cmath> We can isolate <math>\sin x</math> without worrying about division by <math>0</math> since <math>a \neq b \neq 0</math> our final answer is <math>(E) \frac{2ab}{a^2+b^2}</math>

Latest revision as of 18:14, 2 January 2024

Problem 20

If $\tan x=\dfrac{2ab}{a^2-b^2}$ where $a>b>0$ and $0^\circ <x<90^\circ$, then $\sin x$ is equal to

$\textbf{(A) }\frac{a}{b}\qquad \textbf{(B) }\frac{b}{a}\qquad \textbf{(C) }\frac{\sqrt{a^2-b^2}}{2a}\qquad \textbf{(D) }\frac{\sqrt{a^2-b^2}}{2ab}\qquad \textbf{(E) }\frac{2ab}{a^2+b^2}$

Solution

Solution

We start by letting $\tan x = \frac{\sin x}{\cos x}$ so that our equation is now: \[\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}\] Multiplying through and rearranging gives us the equation: \[\cos x = \frac{a^2-b^2}{2ab} * \sin x\] We now apply the Pythagorean identity $\sin ^2 x + \cos ^2 x =1$, using our substitution: \[\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1\] We can isolate $\sin x$ without worrying about division by $0$ since $a \neq b \neq 0$ our final answer is $(E) \frac{2ab}{a^2+b^2}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1972_AHSME_Problems/Problem_20&oldid=209614"