# 1988 AHSME Problems/Problem 13

## Problem

If $\sin(x) =3 \cos(x)$ then what is $\sin(x) \cdot \cos(x)$? $\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{5}\qquad \textbf{(C)}\ \frac{2}{9}\qquad \textbf{(D)}\ \frac{1}{4}\qquad \textbf{(E)}\ \frac{3}{10}$

## Solution

In the problem we are given that $\sin{(x)}=3\cos{(x)}$, and we want to find $\sin{(x)}\cos{(x)}$. We can divide both sides of the original equation by $\cos{(x)}$ to get $$\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.$$ 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 $x$ is acute. If $x$ is obtuse, then $\sin{(x)}$ is positive and $\cos{(x)}$ is negative, so the equation cannot be satisfied. If $x$ is reflex, then both $\sin{(x)}$ and $\cos{(x)}$ are negative, so the equation is satisfied, but when we find $\sin{(x)}\cos{(x)}$, the two negatives will cancel out and give the same (positive) answer as in the acute case.)

Since the problem asks us to find $\sin{(x)}\cos{(x)}$. $$\sin{(x)}\cos{(x)}=\left(\frac{3}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right)=\frac{3}{10}.$$ So $\boxed{\textbf{(E)}\ \frac{3}{10}}$ is our answer.

## Solution 2 (Pure Algebra)

Squaring both sides gives ${\sin}^2 x = 9{\cos}^2 x$. We can take the Pythagorean identity, ${\sin}^2 x + {\cos}^2 x = 1$ and substitute the 1st equation in, giving $10{\cos}^2 x = 1$. So ${\cos}^2 x = \frac{1}{10}$, and ${\sin}^2 x = \frac{9}{10}$.

Multiplying the 2 together gives ${\sin}^2 x {\cos}^2 x = \frac{9}{100}$, and then taking the square root gives $\mp \frac{3}{10}$. However, $-\frac{3}{10}$ implies one of $\sin x$ and $\cos x$ is negative while the other is positive, but $\sin x = 3 \cos x$ means they have the same sign, which contradicts the first statement. This means $\boxed{\textbf{(E)}\ \frac{3}{10}}$ is the only answer.

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