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 "1978 AHSME Problems/Problem 15"

(Created page with "==Solution== Squaring the equation, we get <cmath>\sin ^2 x + \cos ^2 x + 2 \sin x \cos x = \frac{1}{25} =\Rrightarrow 2\sin x \cos x = -\frac{24}{25} \Rrightarrow \sin x \cos...")
 
 
Line 1: Line 1:
 +
== Problem 15 ==
 +
 +
If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is
 +
 +
<math>\textbf{(A) }-\frac{4}{3}\qquad
 +
\textbf{(B) }-\frac{3}{4}\qquad
 +
\textbf{(C) }\frac{3}{4}\qquad
 +
\textbf{(D) }\frac{4}{3}\qquad\\
 +
\textbf{(E) }\text{not completely determined by the given information}    </math>
 +
 
==Solution==
 
==Solution==
 
Squaring the equation, we get
 
Squaring the equation, we get
Line 15: Line 25:
  
 
~JustinLee2017
 
~JustinLee2017
 +
 +
==See Also==
 +
{{AHSME box|year=1978|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 21:10, 13 February 2021

Problem 15

If $\sin x+\cos x=1/5$ and $0\le x<\pi$, then $\tan x$ is

$\textbf{(A) }-\frac{4}{3}\qquad \textbf{(B) }-\frac{3}{4}\qquad \textbf{(C) }\frac{3}{4}\qquad \textbf{(D) }\frac{4}{3}\qquad\\ \textbf{(E) }\text{not completely determined by the given information}$

Solution

Squaring the equation, we get \[\sin ^2 x + \cos ^2 x + 2 \sin x \cos x = \frac{1}{25} =\Rrightarrow 2\sin x \cos x = -\frac{24}{25} \Rrightarrow \sin x \cos x = -\frac{12}{25}\] Recall that $(a-b)^2 = (a+b)^2 - 4ab$, so \[(\sin x - \cos x)^2 = (\sin x + \cos x)^2 - 4 \sin x \cos x\] \[(\sin x - \cos x)^2 = \frac{1}{25} - 4 (- \frac{12}{25})\] \[(\sin x - \cos x) = \frac{7}{5}\] We can now solve for the values of $\sin x$ and $\cos x$ \[\sin x + \cos x = \frac{1}{5}\] \[\sin x - \cos x = \frac{7}{5}\] \[2 \sin x = \frac{8}{5} \Rrightarrow \sin x = \frac{4}{5}, \cos x = -\frac{3}{5}\] Since $\tan x = \frac{\sin x}{\cos x}$, we have \[\tan x = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3}\] $\boxed{A}$

~JustinLee2017

See Also

1978 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_15&oldid=146529"