Ta có
$\cos a = 2\cos^2 (\dfrac{a}{2})-1 = 1 - 2\sin^2(\dfrac{a}{2}) = \cos^2(\dfrac{a}{2}) - \sin^2(\dfrac{a}{2})$
Thật vậy, áp dụng cong thức cos của tổng ta có
$\cos a = \cos(\dfrac{a}{2} + \dfrac{a}{2})$
$= \cos (\dfrac{a}{2}) \cos (\dfrac{a}{2}) - \sin(\dfrac{a}{2}) \sin(\dfrac{a}{2})$
$= \cos^2(\dfrac{a}{2}) - \sin^2(\dfrac{a}{2})$
Mặt khác, ta có
$\cos^2(\dfrac{a}{2}) - \sin^2(\dfrac{a}{2}) = (1-\sin^2(\dfrac{a}{2})) - \sin^2(\dfrac{a}{2})$
$= 1 - 2\sin^2(\dfrac{a}{2})$
$= 1 - 2(1 - \cos^2(\dfrac{a}{2}))$
$= 1 - 2 + 2\cos^2(\dfrac{a}{2})$
$= 2\cos^2(\dfrac{a}{2}) - 1$
