Well we have [tex]\cos 2a = \cos^2 a - \sin^2 a[/tex].
We also know from another well known identity that [tex]\sin^2 a + \cos^2 a =1[/tex] 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 [itex]O^2 + A^2 = H^2[/itex].
If we divide everything by H^2, [tex]\left(\frac{O}{H}\right)^2 +\left(\frac{A}{H}\right)^2 = 1[/tex].
But since O/H is sin a, and A/H is cos a, we have that nice relationship.
Now, since [itex]\sin^2 a + \cos^2 a =1[/tex], we can take either sin^2 a or cos^2 a to the other side: [tex]\cos^2 a = 1 - \sin^2 a[/tex] and [tex]\sin^2 a = 1 - \cos^2 a[/tex].<br />
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Now from the original [tex]\cos^2 a - \sin^2 a[/tex], 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 =][/itex]