Question #fe620

Question

2320 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-25T05:14:12

$cos 2 theta = cos^2 theta - sin ^2 theta$

$tan 2 theta = (2 tan theta)/(1-tan^2 theta)$

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

For example, if $theta = 30^@$
The ratio of a $30^@-60^@-90^@$ triangle is $1: sqrt(3): 2$

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

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

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

$cos 2 theta = cos^2 theta - sin ^2 theta$

$cos 2*30^@ = 3/4 - 1/4 = 2/4 = 1/2$

$tan 2 theta = (2 tan theta)/(1-tan^2 theta)$

$tan 2*30^@ = (2 *1/sqrt(3))/(1-1/3) = (2/sqrt(3))/(2/3)$

$= 2/sqrt(3) * 3/2 = 3/sqrt(3) * sqrt(3)/sqrt(3) = (3 sqrt(3))/3 = sqrt(3)$