Required to Prove the following trig identity:
(cos2x)^2 + (sin2x)^2 = 1
$\displaystyle \cos^2(2x) = (\cos^2{x}-\sin^2{x})^2 = \cos^4{x} - 2\sin^2{x}\cos^2{x} + \sin^4{x}$
$\displaystyle \sin^2(2x) = (2\sin{x}\cos{x})^2 = 4\sin^2{x}\cos^2{x}$
add'em up ...
$\displaystyle \cos^4{x} + 2\sin^2{x}\cos^2{x} + \sin^4{x} = (\cos^2{x} + \sin^2{x})^2 = (1)^2 = 1$
(I don't know if you can prove it this way.)
Draw a unit circle. Plot a point P(x, y) on the unit circle. Drop a perpendicular from point P to the x-axis. You formed a right triangle. Then
$\displaystyle \cos \theta = \frac{\text{adj}}{\text{hyp}} = \frac{x}{1} = x$
and
$\displaystyle \sin \theta = \frac{\text{opp}}{\text{hyp}} = \frac{y}{1} = y$
According to the Pythagorean theorem,
$\displaystyle x^2 + y^2 = 1$,
so plug in $\displaystyle \cos \theta$ for x and $\displaystyle \sin \theta$ for y and you get
$\displaystyle \cos^2 \theta + \sin^2 \theta = 1$.
01
EDIT: Wow, I completely misread the problem. Didn't see the 2 in front of the x.