# How do I show that both sides of these trigonometric identities are equal to each other?

## 1.) $\sin \theta = \cos \theta \tan \theta$ 2.) ${\sin}^{2} \theta - {\cos}^{2} \theta = 1 - 2 {\cos}^{2} \theta$

May 15, 2018

See explanation

#### Explanation:

...pretty straightforward, I think:
First one: $\sin \theta = \cos \theta \tan \theta$
...remember that $\tan \theta = \frac{\sin \theta}{\cos \theta}$
so:
$\sin \theta = \cos \theta \cdot \left(\frac{\sin \theta}{\cos \theta}\right)$

...and the $\cos \theta$ terms on the right cancel out:

$\sin \theta = \sin \theta$

What you have to do for the second one is add $2 {\cos}^{2} \theta$
to both sides:

${\sin}^{2} \theta - {\cos}^{2} \theta + 2 {\cos}^{2} \theta = 1$
${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

...the above is a trig identity you should know by heart, so this shows that the original formula is a valid identity.

GOOD LUCK