# How do you verify 2(tan(2A)) * (2(cos^2(2A) - sin^2(4A)) = sin(8A)?

Apr 15, 2016

shown below

#### Explanation:

$2 \tan \left(2 A\right) \times 2 \left[{\cos}^{2} \left(2 A\right) - {\sin}^{2} \left(4 A\right)\right] = \sin \left(8 A\right)$

LHS=left hand side and RHS=right hand side. So I start with the left hand side and show that it equals the right hand side.

$L H S = 2 \tan \left(2 A\right) \times \left[2 {\cos}^{2} \left(2 A\right) - 2 {\sin}^{2} \left(4 A\right)\right]$

$= 4 \tan \left(2 A\right) {\cos}^{2} \left(2 A\right) - 4 \tan 2 A {\sin}^{2} \left(4 A\right)$

$= 4 \frac{\sin \left(2 A\right)}{\cos} \left(2 A\right) {\cos}^{2} \left(2 A\right) - 4 \frac{\sin \left(2 A\right)}{\cos} \left(2 A\right) {\sin}^{2} \left(4 A\right)$

$= 4 \sin \left(2 A\right) \cos \left(2 A\right) - 4 \frac{\sin \left(2 A\right)}{\cos} \left(2 A\right) {\sin}^{2} \left(2 \left(2 A\right)\right)$

$= 2 \cdot 2 \sin \left(2 A\right) \cos \left(2 A\right) - 4 \frac{\sin \left(2 A\right)}{\cos} \left(2 A\right) \times 2 {\sin}^{2} \left(2 A\right) {\cos}^{2} \left(2 A\right)$

$= 2 \sin \left(2 \left(2 A\right)\right) - 4 \left(\sin \left(2 A\right)\right) \times 2 {\sin}^{2} \left(2 A\right) \cos \left(2 A\right)$

$= 2 \sin \left(4 A\right) - 4 \cdot 2 \sin \left(2 A\right) \cos \left(2 A\right) \times {\sin}^{2} \left(2 A\right)$

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

$= 2 \sin \left(4 A\right) \left[1 - 2 {\sin}^{2} \left(2 A\right)\right]$

$= 2 \sin \left(4 A\right) \cos 2 \left(2 A\right)$

$= 2 \sin \left(4 A\right) \cos \left(4 A\right)$

$= \sin \left(2 \left(4 A\right)\right)$

$= \sin \left(8 A\right)$

$= R H S$