Question

If

Answer 1

Answer:  see proof below

Step-by-step explanation:

Given: cos 330 = $\frac{\sqrt3}{2}$

Use the Double-Angle Identity: cos 2A = 2 cos² A - 1

$\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}$

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:                        $2 \cos^2 165 = \dfrac{\sqrt3}{2} + 1$

                              $\rightarrow 2 \cos^2 165 = \dfrac{\sqrt3}{2} + \dfrac{2}{2}$

Divide by 2:               $\cos^2 165 = \dfrac{\sqrt3+2}{4}$

                             $\rightarrow \cos^2 165 = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}$

                             $\rightarrow \cos^2 165 = \dfrac{2\sqrt3+4}{8}$

Square root:             $\sqrt{\cos^2 165} = \sqrt{\dfrac{4+2\sqrt3}{8}}$

Scratchwork:            $\cos^2 165 = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2$

                             $\rightarrow \cos 165 = \pm \dfrac{\sqrt3+1}{2\sqrt2}$

             Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE

                             $\rightarrow \cos 165 = - \dfrac{\sqrt3+1}{2\sqrt2}$

LHS = RHS $\checkmark$