# Trig Problem

• Mar 4th 2008, 02:31 AM
farso
Trig Problem
Hi there.

Ive had a good look on the forum and cant seem to find anything to help me in this situation, just cant seem to figure this one out, here goes anyway

$\displaystyle \tan 2x = \frac{1}{3}$ where $\displaystyle \tan x = -3 \pm\sqrt{10}$

Any help with this would be really appreciated, ive been looking at it hours now and its just not making sence to me.

Ive had a look at double angle formula, but they dont seem to relate to each other.

Thanks very much
• Mar 4th 2008, 02:51 AM
mr fantastic
Quote:

Originally Posted by farso
Hi there.

Ive had a good look on the forum and cant seem to find anything to help me in this situation, just cant seem to figure this one out, here goes anyway

$\displaystyle \tan 2x = \frac{1}{3}$ where $\displaystyle \tan x = -3 \pm\sqrt{10}$

Any help with this would be really appreciated, ive been looking at it hours now and its just not making sence to me.

Ive had a look at double angle formula, but they dont seem to relate to each other.

Thanks very much

Double angle formula: $\displaystyle \tan(2x) = \frac{2 \tan x}{1 - \tan^2 x}$.

Therefore:

$\displaystyle \frac{1}{3} = \frac{2 \tan x}{1 - \tan^2 x}$

$\displaystyle \Rightarrow \frac{1}{3} (1 - \tan^2 x) = 2 \tan x$

$\displaystyle \Rightarrow 1 - \tan^2 x = 6 \tan x$

$\displaystyle \Rightarrow \tan^2 x + 6 \tan x - 1 = 0$.

This is a quadratic equation in tan x. Solve it using the quadratic formula:

$\displaystyle \tan x = -3 \pm \sqrt{10}$.
• Mar 4th 2008, 03:20 AM
farso
Thanks very much! I can see exactly how i was ment to do that now the answer is in front of me!

Always the way isnt it?

Should be good for next time though, thanks!