# Trigonometric identities.

• Mar 19th 2010, 03:19 AM
somanyquestions
Trigonometric identities.
(sinθ-cosθ)^2
= sin^2 θ-2sinθcosθ-cos^2 θ

is this all?? what else can i do?
• Mar 19th 2010, 03:38 AM
pickslides
Not sure what you are trying to do but

$\displaystyle (\sin\theta-\cos\theta)^2 = (\sin\theta-\cos\theta)(\sin\theta-\cos\theta)= \sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta$ $\displaystyle \neq \sin^2\theta-2\sin\theta\cos\theta-\cos^2\theta$

Also $\displaystyle \cos^2\theta = 1-\sin^2\theta$

and $\displaystyle 2\sin\theta\cos\theta = \sin 2\theta$

Good luck!
• Mar 19th 2010, 07:11 AM
skeeter
Quote:

Originally Posted by somanyquestions
(sinθ-cosθ)^2
= sin^2 θ-2sinθcosθ+cos^2 θ

is this all?? what else can i do?

two identities involved here ...

(1) the basic Pythagorean identity ...

$\displaystyle \sin^2{x} + \cos^2{x} = 1$

(2) the double angle identity for sine ...

$\displaystyle \sin(2x) = 2\sin{x}\cos{x}$

...

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

$\displaystyle 1 - 2\sin{\theta}\cos{\theta} =$

$\displaystyle 1 - \sin(2\theta)$