If $\sin(x)=3\cos(x)$, compute $\sin(x)*\cos(x)$

Question

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?

Answer 1

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.

Answer 2

$$\begin{align} \\ \sin (2x)&=2\sin (x)\cos (x) \\ &=\frac {2\tan (x)}{1+\tan^2 (x)} \\ &=\frac {2\cdot3}{1+(3)^2} \\ &=\frac {6}{10} \\ \end{align}$$

thus

$$\sin (x)\cos (x)=\frac {3}{10}$$

Answer 3

Since $\tan x=3$, $\sin x \cos x = \frac{\tan x}{1+\tan^2 x}$ takes the claimed value.

Answer 4

Hint:

$$ 1=\sin^2(x)+\cos^2(x)=9\cos^2(x)+\cos^2(x). $$

Answer 5

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 $$ \sin x\cos x=\frac{3}{\sqrt{10}}\frac{1}{\sqrt{10}}=\frac{3}{10}. $$

Answer 6

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

$\sin x \cos x = \dfrac{3}{\sqrt{10}} \dfrac{1}{\sqrt{10}} = \dfrac{3}{10}. \tag 9$