Well we have \cos 2a = \cos^2 a - \sin^2 a.
We also know from another well known identity that \sin^2 a + \cos^2 a =1 for all values of a. We can see that because it we have a right angled triangle and label one other angle as a, the adjacent side as A, the opposite side as O and the hypotenuse as H, by Pythagoras O^2 + A^2 = H^2.
If we divide everything by H^2, \left(\frac{O}{H}\right)^2 +\left(\frac{A}{H}\right)^2 = 1.
But since O/H is sin a, and A/H is cos a, we have that nice relationship.
Now, since \sin^2 a + \cos^2 a =1[/tex], we can take either sin^2 a or cos^2 a to the other side: \cos^2 a = 1 - \sin^2 a and \sin^2 a = 1 - \cos^2 a.<br />
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Now from the original \cos^2 a - \sin^2 a, we replace sin^2 a with (1 - cos^2 a), we get the second form, and if we replace cos^2 a with (1 - sin^2 a) we get the third form =]