# How do you prove sin^4 X = (cos^2 2X - 2 cos 2x + 1) / 4?

Apr 30, 2016

Using the following:

• ${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$
• $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$
• $1 = {\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)$

we have:

$\frac{{\cos}^{2} \left(2 x\right) - 2 \cos \left(2 x\right) + 1}{4} = {\left(\cos \left(2 x\right) - 1\right)}^{2} / 4$

$= {\left(\left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right) - \left({\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)\right)\right)}^{2} / 4$

$= {\left(- 2 {\sin}^{2} \left(x\right)\right)}^{2} / 4$

$= \frac{4 {\sin}^{4} \left(x\right)}{4}$

$= {\sin}^{4} \left(x\right)$