Given that cotx = 2/5, then what is the value of tan(2x), cos(2x) and sin(2x)?

Mar 27, 2017

$\tan 2 x = - \frac{20}{21}$
$\sin 2 x \setminus = \frac{20}{29}$
$\cos 2 x \setminus = - \frac{21}{29}$

Explanation:

$\cot x = \frac{2}{5} \implies \tan x = \frac{5}{2}$

Consider the following right angle triangle: By elementary trigonometry we have:

$\tan \theta = \text{opp"/"adj} = \frac{5}{2} \implies \theta = x$

By Pythagoras:

$\setminus \setminus \setminus \setminus {h}^{2} = {2}^{2} + {5}^{2}$
$\therefore {h}^{2} = 4 + 25$
$\therefore {h}^{2} = 29$
$\therefore \setminus \setminus h = \sqrt{29}$

And so;

$\sin x = \text{opp"/"hyp} = \frac{5}{h} = \frac{5}{\sqrt{29}}$
$\cos x = \text{adj"/"hyp} = \frac{2}{h} = \frac{2}{\sqrt{29}}$

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

$\tan 2 x = \frac{2 \tan x}{1 - {\tan}^{2} x}$
$\text{ } = \frac{2 \cdot \frac{5}{2}}{1 - {\left(\frac{5}{2}\right)}^{2}}$
$\text{ } = \frac{5}{1 - \frac{25}{4}}$
$\text{ } = \frac{5}{- \frac{21}{4}}$
$\text{ } = - \frac{20}{21}$

And similarly using $\sin 2 x = 2 \sin x \cos x$

$\sin 2 x = 2 \cdot \frac{5}{\sqrt{29}} \cdot \frac{2}{\sqrt{29}}$
$\text{ } = \frac{20}{29}$

And similarly using $\cos 2 x = {\cos}^{2} x - {\sin}^{2} x$

$\cos 2 x = {\left(\frac{2}{\sqrt{29}}\right)}^{2} - {\left(\frac{5}{\sqrt{29}}\right)}^{2}$
$\text{ } = \frac{4}{29} - \frac{25}{29}$
$\text{ } = - \frac{21}{29}$