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

Prove:

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

Any help would be great.

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- Apr 2nd 2009, 06:10 AMDominic O'LProve the following Trig Identity
Sorry for troubling you guys but i keep coming up with an extra 3cos

**X**

**Prove:**

3sin**x**tan**x=**8*is the same as*3cos^2**x**+8cos**x**-3=0

Any help would be great. - Apr 2nd 2009, 06:38 AMe^(i*pi)
$\displaystyle tan{x} = \frac{sin(x)}{cos(x)}$

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

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

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

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

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

Then cos(x) will disappear and take everything to the other side to get the answer - Apr 2nd 2009, 06:42 AMmasters
Hi Dominic,

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

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

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

$\displaystyle 3\cos x-1=0$

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

Now, use that in the 1st equation.

$\displaystyle 3 \sin x \tan x = 8$

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

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

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

$\displaystyle 9 \sin^2x=8$

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

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

Therefore, $\displaystyle \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)=\cos^{-1} \left(\frac{1}{3}\right)$ - Apr 2nd 2009, 06:50 AMDominic O'L
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(Worried).

- Apr 2nd 2009, 07:56 AMSoroban
Hello, Dominic O'L

Quote:

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

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

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

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