1. ## Trig problem

Hi. Got a maths trig question.
For a angle of less than 90 degrees, sinx=3cosx. Show and decide the value of sin2x.

I just cant bring myself to figure this one out. Helps or pointers would be very appreciated

2. Originally Posted by Burger king
Hi. Got a maths trig question.
For a angle of less than 90 degrees, sinx=3cosx. Show and decide the value of sin2x.

I just cant bring myself to figure this one out. Helps or pointers would be very appreciated
$\sin(2x) = 2\sin{x}\cos{x}$

since $\sin{x} = 3\cos{x}$ ...

$
\sin(2x) = 2(3\cos{x})\cos{x} = 6\cos^2{x}
$

3. Hello, Burger king!

I can't imagine what "show and decide" means.

For an angle $x < 90^o\!:\;\;\sin x\:=\:3\cos x.$
Show and decide the value of $\sin2x.$

We have: . $\sin x \:=\:3\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:=\:3 \quad\Rightarrow\quad \tan x \:=\:3$
Hence: . $\tan x \:=\:\frac{3}{1} \:=\:\frac{opp}{adj}$

So $x$ is an angle in a right triangle with: $opp = 3,\;adj = 1$
. . Pythagorus says: . $hyp = \sqrt{10}$
Hence: . $\sin x = \tfrac{3}{\sqrt{10}},\;\;\cos x = \tfrac{1}{\sqrt{10}}$

Therefore: . $\sin2x \:=\:2\sin x\cos x \:=\:2\left(\frac{3}{\sqrt{10}}\right)\left(\frac{ 1}{\sqrt{10}}\right) \:=\:\frac{3}{5}$