Answer: see proof below
Step-by-step explanation:
Use the following Sum/Difference Identities:
sin(A + B) = sin A · cos B + sin B · cos A
sin(A - B) = sin A · cos B - sin B · cos A
Use the following Half-Angle Identities:
[tex]\sin\bigg(\dfrac{\theta}{2}\bigg)=\dfrac{\sqrt{1-\cos \theta}}{\sqrt2}\\\\\\\cos\bigg(\dfrac{\theta}{2}\bigg)=\dfrac{\sqrt{1+\cos \theta}}{\sqrt2}[/tex]
[tex]\text{Use the Unit circle to evaluate:}\ \cos\dfrac{\pi}{4} = \sin\dfrac{\pi}{4} = \dfrac{\sqrt2}{2}[/tex]
Use the following side work:
[tex]\sin\bigg(\dfrac{\pi}{8}\bigg)=\sin\bigg(\dfrac{\frac{\pi}{4}}{2}\bigg)=\dfrac{\sqrt{1-\cos \frac{\pi}{4}}}{\sqrt2}=\dfrac{\sqrt{2-\sqrt2}}{2}\\\\\\\cos\bigg(\dfrac{\pi}{8}\bigg)=\cos\bigg(\dfrac{\frac{\pi}{4}}{2}\bigg)=\dfrac{\sqrt{1+\cos \frac{\pi}{4}}}{\sqrt2}=\dfrac{\sqrt{2+\sqrt2}}{2}[/tex]
Proof LHS → RHS
[tex]\text{LHS:}\qquad \qquad \qquad \qquad \qquad \sin^2\bigg(\dfrac{\pi}{8}+\dfrac{A}{2}\bigg)-\sin^2\bigg(\dfrac{\pi}{8}-\dfrac{A}{2}\bigg)\\\\\text{Sum/Difference Identity:}\qquad \bigg(\sin\dfrac{\pi}{8}\cdot \cos \dfrac{A}{2}+\sin \dfrac{A}{2}\cdot \cos \dfrac{\pi}{8}\bigg)^2\\\\.\qquad \qquad \qquad\qquad \qquad \quad -\bigg(\sin\dfrac{\pi}{8}\cdot \cos \dfrac{A}{2}-\sin \dfrac{A}{2}\cdot \cos \dfrac{\pi}{8}\bigg)^2[/tex]
[tex]\text{Expand and Simplify:}\qquad \quad 4\sin \dfrac{\pi}{8}\cdot \cos\dfrac{A}{2}\cdot \sin \dfrac{A}{2}\cdot \cos \dfrac{A}{2}\\\\\\\text{Substitute:}\qquad \qquad \qquad 4\bigg(\dfrac{\sqrt{2-\sqrt2}}{2}\bigg)\cdot \cos \dfrac{A}{2}\cdot \sin \dfrac{A}{2}\bigg(\dfrac{\sqrt{2+\sqrt2}}{2}\bigg)\\\\\\\text{Simplify:}\qquad \qquad \qquad \sqrt2\cos \bigg(\dfrac{A}{2}\bigg)\cdot \sin \bigg(\dfrac{A}{2}\bigg)[/tex]
[tex]\text{Half-Angle Identity:} \quad \sqrt2\bigg(\dfrac{\sqrt{1+\cos A}}{\sqrt2}\bigg)\bigg(\dfrac{\sqrt{1-\cos A}}{\sqrt2}\bigg)\\\\\\\text{Simplify:}\qquad \qquad \qquad \dfrac{\sqrt{1-\cos^2 A}}{\sqrt2}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{\sqrt{\sin^2 A}}{\sqrt2}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{\sin A}{\sqrt2}[/tex]
[tex]\text{LHS = RHS:}\quad \dfrac{\sin A}{\sqrt2} = \dfrac{\sin A}{\sqrt2}\quad \checkmark[/tex]
