Answer: see proof below
Step-by-step explanation:
Given: cos 330 = [tex]\frac{\sqrt3}{2}[/tex]
Use the Double-Angle Identity: cos 2A = 2 cos² A - 1
[tex]\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}[/tex]
Proof LHS → RHS:
LHS cos 165
Double-Angle: cos (2 · 165) = 2 cos² 165 - 1
⇒ cos 330 = 2 cos² 165 - 1
⇒ 2 cos² 165 = cos 330 + 1
Given: [tex]2 \cos^2 165 = \dfrac{\sqrt3}{2} + 1[/tex]
[tex]\rightarrow 2 \cos^2 165 = \dfrac{\sqrt3}{2} + \dfrac{2}{2}[/tex]
Divide by 2: [tex]\cos^2 165 = \dfrac{\sqrt3+2}{4}[/tex]
[tex]\rightarrow \cos^2 165 = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}[/tex]
[tex]\rightarrow \cos^2 165 = \dfrac{2\sqrt3+4}{8}[/tex]
Square root: [tex]\sqrt{\cos^2 165} = \sqrt{\dfrac{4+2\sqrt3}{8}}[/tex]
Scratchwork: [tex]\cos^2 165 = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2[/tex]
[tex]\rightarrow \cos 165 = \pm \dfrac{\sqrt3+1}{2\sqrt2}[/tex]
Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE
[tex]\rightarrow \cos 165 = - \dfrac{\sqrt3+1}{2\sqrt2}[/tex]
LHS = RHS [tex]\checkmark[/tex]
