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Difference between revisions of "1988 AHSME Problems/Problem 13"
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Squaring both sides gives <math>{\sin}^2 x = 9{\cos}^2 x</math>. We can take the Pythagorean identity, <math>{\sin}^2 x + {\cos}^2 x = 1</math> and substitute the 1st equation in, giving <math>10{\cos}^2 x = 1</math>. So <math>{\cos}^2 x = \frac{1}{10}</math>, and <math>{\sin}^2 x = \frac{9}{10}</math>.  
 
Squaring both sides gives <math>{\sin}^2 x = 9{\cos}^2 x</math>. We can take the Pythagorean identity, <math>{\sin}^2 x + {\cos}^2 x = 1</math> and substitute the 1st equation in, giving <math>10{\cos}^2 x = 1</math>. So <math>{\cos}^2 x = \frac{1}{10}</math>, and <math>{\sin}^2 x = \frac{9}{10}</math>.  
  
−
Multiplying the 2 together gives <math>{\sin}^2{\cos}^2 x = \frac{9}{100}</math>, and then taking the square root gives <math>\frac{3}{10}</math>. Keep in mind <math>-\frac{3}{10}</math> implies one of <math>\sin x</math> and <math>\cos x</math> is negative while the other is positive, but <math>\sin x = 3 \cos x</math> means they have the same sign.
+
Multiplying the 2 together gives <math>{\sin}^2 x {\cos}^2 x = \frac{9}{100}</math>, and then taking the square root gives <math>\mp \frac{3}{10}</math>. However, <math>-\frac{3}{10}</math> implies one of <math>\sin x</math> and <math>\cos x</math> is negative while the other is positive, but <math>\sin x = 3 \cos x</math> means they have the same sign, which contradicts the first statement. This means <math>\boxed{\textbf{(E)}\ \frac{3}{10}}</math> is the only answer.
  
 
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-ThisUsernameIsTaken

Latest revision as of 19:53, 24 August 2021

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.

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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

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