If $\sin(x)=3\cos(x)$, compute $\sin(x)*\cos(x)$ I drew a triangle and was instructed to use a property unknown to me. Apparently the answer is $\frac{3}{\sqrt{10}}*\frac{1}{\sqrt{10}}=\frac{3}{10}$. Is this answer correct, and what theorem/formula is this?

We have the identity
$1 + \tan^2 \theta = \sec^2 \theta = \dfrac{1}{\cos^2 \theta}; \tag 1$
so with
$\sin x = 3\cos x, \tag 2$
we infer
$\tan x = \dfrac{\sin x}{\cos x} = 3, \tag 3$
whence
$\dfrac{1}{\cos^2 x} = 1 + \tan^2 x = 1 + 3^2 = 10, \tag 4$
yielding
$\cos x = \pm \dfrac{1}{\sqrt{10}}; \tag 5$
since
$\sin^2 x + \cos^2 x = 1, \tag 6$
we infer
$\sin^2 x = 1 - \dfrac{1}{10} = \dfrac{9}{10}, \tag 7$
whence
$\sin x = \pm \dfrac{3}{\sqrt{10}}; \tag 8$
now, as has been pointed out by Foobaz John in his answer,  $\tan x = 3$ implies the sign of $\sin x$ and $\cos x$ must be the same; thus 
$\sin x \cos x = \dfrac{3}{\sqrt{10}} \dfrac{1}{\sqrt{10}} = \dfrac{3}{10}. \tag 9$
