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?
 A: $$\begin{align} \\
\sin (2x)&=2\sin (x)\cos (x) \\
&=\frac {2\tan (x)}{1+\tan^2 (x)} \\
&=\frac {2\cdot3}{1+(3)^2} \\
&=\frac {6}{10} \\
\end{align}$$
thus
$$\sin (x)\cos (x)=\frac {3}{10}$$
A: Since $\tan x=3$, $\sin x \cos x = \frac{\tan x}{1+\tan^2 x}$ takes the claimed value.
A: Observe that
$$
\tan (x)=3.
$$
But
$$
1+\tan^2x=\sec^2x\implies10=\frac{1}{\cos^2x}
$$
and
$$
1+\cot^2x=\csc^2x
\implies\frac{10}{9}=\frac{1}{\sin^2x}$$
Since tangent is positive, both sine and cosine have the same sign. In particular
$$
\sin x\cos x=\frac{3}{\sqrt{10}}\frac{1}{\sqrt{10}}=\frac{3}{10}.
$$
A: We have $\tan(x) = 3$
Draw a right angle triangle where $x$ is one of the angle, let the opposite side be of length $3$, let the adjacent side be of length one. Then by Pythagoras Theorem, the hypothenus side is $\sqrt{10}$.
Hence you can compute $\sin(x)$ and $\cos(x)$ and multiply them.
A: 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 
$\sin x \cos x = \dfrac{3}{\sqrt{10}} \dfrac{1}{\sqrt{10}} = \dfrac{3}{10}. \tag 9$
A: Hint:
$$
1=\sin^2(x)+\cos^2(x)=9\cos^2(x)+\cos^2(x).
$$
