Difference between revisions of "1988 AHSME Problems/Problem 13"
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==Problem== | ==Problem== | ||
| − | If <math>\sin(x) =3 \cos(x) </math> then what is <math>\sin(x)\cos(x)</math>? | + | If <math>\sin(x) =3 \cos(x) </math> then what is <math>\sin(x) \cdot \cos(x)</math>? |
<math>\textbf{(A)}\ \frac{1}{6}\qquad | <math>\textbf{(A)}\ \frac{1}{6}\qquad | ||
| Line 7: | Line 7: | ||
\textbf{(C)}\ \frac{2}{9}\qquad | \textbf{(C)}\ \frac{2}{9}\qquad | ||
\textbf{(D)}\ \frac{1}{4}\qquad | \textbf{(D)}\ \frac{1}{4}\qquad | ||
| − | \textbf{(E)}\ \frac{3}{10} </math> | + | \textbf{(E)}\ \frac{3}{10} </math> |
==Solution== | ==Solution== | ||
| + | In the problem we are given that <math>\sin{(x)}=3\cos{(x)}</math>, and we want to find <math>\sin{(x)}\cos{(x)}</math>. We can divide both sides of the original equation by <math>\cos{(x)}</math> to get <cmath>\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.</cmath> | ||
| + | We can now use right triangle trigonometry to finish the problem. | ||
| + | <asy> | ||
| + | pair A,B,C; | ||
| + | A = (0,0); | ||
| + | B = (3,0); | ||
| + | C = (0,1); | ||
| + | draw(A--B--C--A); | ||
| + | draw(rightanglemark(B,A,C,8)); | ||
| + | label("$A$",A,SW); | ||
| + | label("$B$",B,SE); | ||
| + | label("$C$",C,N); | ||
| + | label("$3$",B/2,S); | ||
| + | label("$1$",C/2,W); | ||
| + | label("$\sqrt{10}$",(C+B)/2,NE); | ||
| + | </asy> | ||
| + | (Note that this assumes that <math>x</math> is acute. If <math>x</math> is obtuse, then <math>\sin{(x)}</math> is positive and <math>\cos{(x)}</math> is negative, so the equation cannot be satisfied. If <math>x</math> is reflex, then both <math>\sin{(x)}</math> and <math>\cos{(x)}</math> are negative, so the equation is satisfied, but when we find <math>\sin{(x)}\cos{(x)}</math>, the two negatives will cancel out and give the same (positive) answer as in the acute case.) | ||
| + | Since the problem asks us to find <math>\sin{(x)}\cos{(x)}</math>. | ||
| + | <cmath>\sin{(x)}\cos{(x)}=\left(\frac{3}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right)=\frac{3}{10}.</cmath> | ||
| + | So <math>\boxed{\textbf{(E)}\ \frac{3}{10}}</math> is our answer. | ||
== See also == | == See also == | ||
Revision as of 18:20, 26 February 2018
Problem
If 
then what is 
?
Solution
In the problem we are given that 
, and we want to find 
. We can divide both sides of the original equation by 
to get ![\[\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.\]](http://latex.artofproblemsolving.com/d/c/8/dc885d8a953b450b93c19a12d44ad499db3d198d.png)
We can now use right triangle trigonometry to finish the problem.
![[asy] pair A,B,C; A = (0,0); B = (3,0); C = (0,1); draw(A--B--C--A); draw(rightanglemark(B,A,C,8)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$3$",B/2,S); label("$1$",C/2,W); label("$\sqrt{10}$",(C+B)/2,NE); [/asy]](http://latex.artofproblemsolving.com/f/7/9/f79fa3de0726b6a6cc5f8ebee031fa281fe7fd25.png)
(Note that this assumes that 
is acute. If is obtuse, then 
is positive and is negative, so the equation cannot be satisfied. If is reflex, then both and are negative, so the equation is satisfied, but when we find , the two negatives will cancel out and give the same (positive) answer as in the acute case.)
Since the problem asks us to find .
![\[\sin{(x)}\cos{(x)}=\left(\frac{3}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right)=\frac{3}{10}.\]](http://latex.artofproblemsolving.com/7/d/0/7d09b519c970a94629daae9422c367681b9ecb68.png)
So 
is our answer.
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
