Question

Prove that: ${\sin }^2(\frac{\pi }{8}+\frac{A}{2})-{\sin }^2(\frac{\pi }{8}-\frac{A}{2})=\frac{1}{\sqrt{2}}\sin A$
Find the general solution of the equation.

Answer 1

REF.Image.

OA = x

OB = 180 m

$tan\varphi =4/3$

$sin\varphi =4/5$

$sin\varphi =OA/Oϵ$

$sin\varphi =\frac{x}{180}$

$4/5=x/180$

$x=144$

$144+10=154m$

$sin(\pi /8+A/2)-si{n}^2(\pi /8-\frac{A}{2})=\frac{1}{\sqrt{2}}sinA$

$sin\left(\pi /4\right)sin\left(A\right)=\frac{1}{\sqrt{2}}sinA$

$Aϵ\left(2n\pi \right)ϵR$