help please! i think the pathagoream[sp*] theorem is used here but i don't know where.
This is not really an a problem about solve for x,because any makes it true! This is called an identity problem.Originally Posted by CONFUSED_ONE
You need to demonstrate.
$\displaystyle \cos^2x-\sin^2x=2\cos^2x-1$
By Pythagorean Identity you have that,
$\displaystyle \sin^2x=1-\cos^2x$
Thus, (watch those signs )
$\displaystyle \cos^2x-(1-\cos^2x)=2\cos^2x-1$
Thus, we have on Left Hand Side,
$\displaystyle 2\cos^2x-1=2\cos^2x-1$
(Note I change the name of this thread for reasons explained above).
$\displaystyle \mathbb{Q}.\mathbb{E}.\mathbb{D}$