Answer: see proof below
Step-by-step explanation:
Use the following Double Angle Identity:
cos 2A = 2 cos²A - 1
Proof RHS → LHS
Given: [tex]\sqrt{2+\sqrt{2+\sqrt{2+2cos8A}}}[/tex]
Factor: [tex]\sqrt{2+\sqrt{2+\sqrt{2(1+cos8A)}}}[/tex]
Let α = 4A: [tex]\sqrt{2+\sqrt{2+\sqrt{2(1+cos2\alpha)}}}[/tex]
Double Angle Identity: [tex]\sqrt{2+\sqrt{2+\sqrt{2(1+2cos^2\alpha-1)}}}[/tex]
Simplify: [tex]\sqrt{2+\sqrt{2+\sqrt{2(2cos^2\alpha)}}}[/tex]
[tex]\sqrt{2+\sqrt{2+2cos\alpha}}}[/tex]
Substitute (α = 4A): [tex]\sqrt{2+\sqrt{2+2cos4A}}}[/tex]
Factor: [tex]\sqrt{2+\sqrt{2(1+cos4A)}}}[/tex]
Let β = 2A [tex]\sqrt{2+\sqrt{2(1+cos2\beta)}}}[/tex]
Double Angle Identity: [tex]\sqrt{2+\sqrt{2(1+2cos^2\beta-1)}}}[/tex]
Simplify: [tex]\sqrt{2+\sqrt{2(2cos^2\beta)}}}[/tex]
[tex]\sqrt{2+2cos\beta}[/tex]
Substitute (β = 2A): [tex]\sqrt{2+2cos2\alpha}[/tex]
Factor: [tex]\sqrt{2(1+cos2\alpha)}[/tex]
Double Angle Identity: [tex]\sqrt{2(1+2cos^2\alpha-1)}[/tex]
Simplify: [tex]\sqrt{2(2cos^2A)}[/tex]
2 cos A
2cos A = 2cos A [tex]\checkmark[/tex]
