# Math Help - Prove the following Trig Identity

1. ## Prove the following Trig Identity

Sorry for troubling you guys but i keep coming up with an extra 3cosX

Prove:

3sinxtanx=8 is the same as 3cos^2x+8cosx-3=0

Any help would be great.

2. Originally Posted by Dominic O'L
Sorry for troubling you guys but i keep coming up with an extra 3cosX

Prove:

3sinxtanx=8 is the same as 3cos^2x+8cosx-3=0

Any help would be great.
$tan{x} = \frac{sin(x)}{cos(x)}$

$\frac{3sin^2(x)}{cos(x)}-8 = 0$

$\frac{3-3cos^2(x)}{cos(x)}-8=0$

$-8 = -8\frac{cos(x)}{cos(x)}$

$\frac{3-3cos^2(x)}{cos(x)}-8\frac{cos(x)}{cos(x)}=0$

$\frac{3-3cos^2(x)-8cos(x)}{cos(x)} = 0$

Then cos(x) will disappear and take everything to the other side to get the answer

3. Originally Posted by Dominic O'L
Sorry for troubling you guys but i keep coming up with an extra 3cosX

Prove:

3sinxtanx=8 is the same as3cos^2x+8cosx-3=0

Any help would be great.
Hi Dominic,

I'm not sure if this is the proof you are looking for, but here is something for what it's worth.

$3 \cos^2x+8 \cos x -3=0$

$(3 \cos x-1)(\cos x +3)=0$

$3\cos x-1=0$

$\cos x={\color{red}\frac{1}{3}}$

Now, use that in the 1st equation.

$3 \sin x \tan x = 8$

$3 \sin x \left(\frac{\sin x}{\cos x}\right)=8$

$\frac{3 \sin^2x}{\cos x}=8$

$\frac{3 \sin^2 x}{{\color{red}\frac{1}{3}}}=8$

$9 \sin^2x=8$

$\sin^2x=\frac{8}{9}$

$\sin x=\frac{2\sqrt{2}}{3}$

Therefore, $\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)=\cos^{-1} \left(\frac{1}{3}\right)$

4. Thank you ever so much, i'm in the process of working through maths past papers so your help is much appreciated, AS exam only a month and a half away.

5. Hello, Dominic O'L

Prove: . $3\sin x\tan x\:=\:8$ is the same as $3\cos^2\!x+ 8\cos x -3\:=\:0$

We have: . $3\sin x\cdot\frac{\sin x}{\cos x} \:=\:8 \quad\Rightarrow\quad \frac{3\sin^2\!x}{\cos x} \:=\:8 \quad\Rightarrow\quad 3\sin^2\!x \:=\:8\cos x$

. . $3(1-\cos^2\!x) \:=\:8\cos x \quad\Rightarrow\quad 3-3\cos^2\!x \:=\: 8\cos x$

Therefore: . $\boxed{3\cos^2\!x + 8\cos x - 3 \:=\:0}$