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
$\displaystyle \sin(2x) = 2\sin{x}\cos{x}$

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

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

3. Hello, Burger king!

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

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

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

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

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