# If cos x=11/12 then find the value of tan2x?

Mar 23, 2017

$\tan 2 x = = \frac{11 \sqrt{23}}{49}$

#### Explanation:

Consider the following right angle triangle:

By elementary trigonometry we have:

$\cos \theta = \text{adj"/"hyp} = \frac{11}{12} \implies \theta = x$

By Pythagoras:

$\setminus \setminus \setminus {12}^{2} = {11}^{2} + {h}^{2}$
$\therefore {h}^{2} = 144 - 121$
$\therefore {h}^{2} = 23$
$\therefore \setminus \setminus h = \sqrt{23}$

And so;

$\tan x = \text{opp"/"adj} = \frac{h}{11} = \frac{\sqrt{23}}{11}$

Using the identity $\tan \left(2 \theta\right) \equiv \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$ we have:
$\tan \left(2 x\right) = \frac{2 \tan x}{1 - {\tan}^{2} x}$

$\text{ } = \frac{2 \frac{\sqrt{23}}{11}}{1 - {\left(\frac{\sqrt{23}}{11}\right)}^{2}}$

$\text{ } = \frac{2 \frac{\sqrt{23}}{11}}{1 - \frac{23}{121}}$

$\text{ } = \frac{2 \frac{\sqrt{23}}{11}}{\frac{98}{121}}$

$\text{ } = \frac{2 \sqrt{23}}{11} \cdot \frac{121}{98}$

$\text{ } = \frac{11 \sqrt{23}}{49}$

For comparison, to verify the solution, if we use a calculator:

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \cos x = \frac{11}{12} \implies x = {23.556}^{o}$
$\therefore \tan 2 x = 1.0766$; and $\frac{11 \sqrt{23}}{49} = 1.0766$

Mar 23, 2017

$\tan \left(2 x\right) = \pm \frac{11 \sqrt{23}}{49}$

#### Explanation:

$\cos x = \frac{11}{12}$

Use the identity $\cos 2 x = 2 {\cos}^{2} x - 1$:

$\cos 2 x = 2 \cdot \frac{121}{144} - 1 = \frac{242 - 144}{144} = \frac{98}{144}$

Use now the identity:

${\tan}^{2} \left(2 x\right) = {\sin}^{2} \frac{2 x}{\cos} ^ 2 \left(2 x\right) = \frac{1 - {\cos}^{2} \left(2 x\right)}{\cos} ^ 2 \left(2 x\right) = \frac{1}{\cos} ^ 2 \left(2 x\right) - 1$

so:

${\tan}^{2} \left(2 x\right) = {\left(144\right)}^{2} / {\left(98\right)}^{2} - 1 = \frac{20736 - 9604}{9604} = \frac{11132}{9604}$

Then:

$\tan \left(2 x\right) = \pm \sqrt{\frac{11132}{9604}} = \pm \frac{\sqrt{2 \cdot 2 \cdot 11 \cdot 11 \cdot 23}}{98} = \pm 22 \frac{\sqrt{23}}{98} = \pm \frac{11}{49} \sqrt{23}$