Question fe620

Apr 25, 2017

$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta$

$\tan 2 \theta = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$

Explanation:

You need to know what the angle $\theta$ is.

For example, if $\theta = {30}^{\circ}$
The ratio of a ${30}^{\circ} - {60}^{\circ} - {90}^{\circ}$ triangle is $1 : \sqrt{3} : 2$

So sin 30^@ = 1/2; sin^2 30^@ = (1/2)^2 = 1/4

cos 30^@ = sqrt(3)/2; cos^2 30^@ = (sqrt(3)/2)^2 = 3/4

tan 30^@ = 1/sqrt(3); tan^2 30^@ = (1/sqrt(3))^2 = 1/3#

Use the trigonometric double angle identities:

$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta$

$\cos 2 \cdot {30}^{\circ} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

$\tan 2 \theta = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$

$\tan 2 \cdot {30}^{\circ} = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

$= \frac{2}{\sqrt{3}} \cdot \frac{3}{2} = \frac{3}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3 \sqrt{3}}{3} = \sqrt{3}$