Simple AHSME (AIME) problem

$\sin{x}=3\cos{x}$ $\frac{\sin{x}}{\cos{x}}=3$ $\tan{x}=3$ So we have a triangle whose base is 3, the height is 1 and is perpendicular to the base, and the hypotenuse is $h$ , which we find to be $\sqrt{10}$ through Pythagorean Theorem. Of course, you could have gotten this simply from the ratio of $\sin{x}$ to $\cos{x}$ , but I found it helpful to solve it out this way.

From here, you know that $\sin{x}=\frac{3}{\sqrt{10}}$ and $\cos{x}=\frac{1}{\sqrt{10}}$ . Therefore, $(\sin{x})(\cos{x})=(\frac{3}{\sqrt{10}})(\frac{1}{\sqrt{10}})=\frac{3}{10}$

(Note: The base is actually $3x$ , the height is $1x$ , and the hypotenuse is $hx$ , where $x$ is any positive real number. However, you'll get the same answer for whatever $x$ you choose because the $x$ 's cancel out, so for the sake of simplicity, I set $x=1$ ).

Given that $\sin x = 3 \cos x$ , $\implies \dfrac {\sin x}{\cos x} = \tan x = 3$ . Then we have:

$\begin{aligned} \sin x \cos x & = \frac 12 \color{#D61F06} \sin 2x & \small \color{#3D99F6} \text{Since }\sin 2\theta = 2 \sin \theta \cos \theta \\ & = \frac 12 \times \color{#D61F06} \frac {2\tan x}{1+\tan^2 x} & \small \color{#D61F06} \text{By half-angle tangent substitution (see reference)} \\ & = \frac 12 \times \frac {2 \times 3}{1+3^2} \\ & = \frac 3{10} = \boxed{0.3} \end{aligned}$

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Reference: Half-angle tangent substitution

• Since sin x = 3cos x, that means the side opposite angle x is 3 times greater than the side adjacent to angle x. Lets call the adjacent side Y and the opposite side 3Y. The hypotenuse will be Y * sqrt10. So (sin x)(cos x) = 0.3.

Tanx equals to 3 ,then sinx into cosx equals to 3/√10 *1√10 ,so,3/10 equals to 0.3