Question

Please solve this question.​

Answer 1

Answer:  see proof below

Step-by-step explanation:

Use the Product to Sum Identity:  cos A · cos B = [cos (A - B) + cos (A + B)]/2

Use the Odd Function Identity: cos (-A) = - cos (A)

Proof LHS → RHS:

$\text{LHS:}\qquad \qquad 2\cos \bigg(\dfrac{11\pi}{16}\bigg)\cdot \cos \bigg(\dfrac{\pi}{16}\bigg)+\cos \bigg(\dfrac{\pi}{4}\bigg)+\cos \bigg(\dfrac{3\pi}{8}\bigg)$

$\text{Prod to Sum:}\quad \dfrac{2\bigg[\cos \bigg(\frac{(11\pi -\pi)}{16}\bigg)+\cos \bigg(\frac{(11\pi+\pi}{16}\bigg)\bigg]}{2}+\cos \bigg(\dfrac{\pi}{4}\bigg)+\cos \bigg(\dfrac{3\pi}{8}\bigg)$

                   $=\cos \bigg(\dfrac{10\pi}{16}\bigg)+\cos \bigg(\dfrac{12\pi}{16}\bigg)+\cos \bigg(\dfrac{\pi}{4}\bigg)+\cos \bigg(\dfrac{3\pi}{8}\bigg)$

                   $=\cos \bigg(\dfrac{5\pi}{8}\bigg)+\cos \bigg(\dfrac{3\pi}{4}\bigg)+\cos \bigg(\dfrac{\pi}{4}\bigg)+\cos \bigg(\dfrac{3\pi}{8}\bigg)$

                   $=\cos \bigg(\dfrac{-3\pi}{8}\bigg)+\cos \bigg(\dfrac{-\pi}{4}\bigg)+\cos \bigg(\dfrac{\pi}{4}\bigg)+\cos \bigg(\dfrac{3\pi}{8}\bigg)$

$\text{Odd Functions:}\quad -\cos \bigg(\dfrac{3\pi}{8}\bigg)-\cos \bigg(\dfrac{\pi}{4}\bigg)+\cos \bigg(\dfrac{\pi}{4}\bigg)+\cos \bigg(\dfrac{3\pi}{8}\bigg)$

                          = 0

LHS = RHS   $\checkmark$