Answer: see proof below
Step-by-step explanation:
Use the following identities:
2cos x · cos y = cos(x + y) + cos(x - y)
cos x + cos y = 2 cos (x + y)/2 · cos(x - y)/2
Use the Unit Circle to evaluate: cos(π/2) = 0
Proof LHS → RHS
[tex]\text{LHS:}\qquad \qquad \qquad 2\cos \dfrac{9\pi}{13}\cos \dfrac{\pi}{13}\quad +\quad \cos \dfrac{3\pi}{13}+\cos \dfrac{5\pi}{13}\\\\\text{Identity:}\qquad \quad \cos\bigg(\dfrac{9\pi}{13}+\dfrac{\pi}{13}\bigg)+ \cos\bigg(\dfrac{9\pi}{13}-\dfrac{\pi}{13}\bigg)+\quad \cos\bigg(\dfrac{3\pi}{13}\bigg)+\cos \bigg(\dfrac{5\pi}{13}\bigg)[/tex]
[tex]\text{Simplify:}\qquad \qquad \cos\bigg( \dfrac{10\pi}{13}\bigg)+\cos \bigg(\dfrac{8\pi}{13}\bigg)+\qquad \cos\bigg(\dfrac{3\pi}{13}\bigg)+\cos \bigg(\dfrac{5\pi}{13}\bigg)[/tex]
[tex]\text{Regroup:}\qquad \qquad \cos\bigg( \dfrac{10\pi}{13}\bigg)+\cos \bigg(\dfrac{3\pi}{13}\bigg)\quad +\quad \cos\bigg(\dfrac{8\pi}{13}\bigg)+\cos \bigg(\dfrac{5\pi}{13}\bigg)[/tex]
[tex]\text{Identity:}\qquad 2\cos \bigg(\dfrac{10\pi+3\pi}{13\cdot 2}\bigg)\cdot \cos \bigg(\dfrac{10\pi-3\pi}{13\cdot 2}\bigg)+\quad 2\cos\bigg(\dfrac{8\pi+5\pi}{13\cdot 2}\bigg)\cdot \cos\bigg(\dfrac{8\pi-5\pi}{13\cdot 2}\bigg)[/tex]
[tex]\text{Simplify:}\qquad 2\cos \bigg(\dfrac{13\pi}{26}\bigg)\cdot \cos \bigg(\dfrac{7\pi}{26}\bigg)+\quad 2\cos\bigg(\dfrac{13\pi}{26}\bigg)\cdot \cos\bigg(\dfrac{3\pi}{26}\bigg)\\\\\\.\qquad \qquad =2\cos \bigg(\dfrac{\pi}{2}\bigg)\cdot \cos \bigg(\dfrac{7\pi}{26}\bigg)+\quad 2\cos\bigg(\dfrac{\pi}{2}\bigg)\cdot \cos\bigg(\dfrac{3\pi}{26}\bigg)\\\\\\\text{Factor:}\qquad =2\cos\bigg(\dfrac{\pi}{2}\bigg)\bigg[ \cos \bigg(\dfrac{7\pi}{26}\bigg)+ \cos \bigg(\dfrac{3\pi}{26}\bigg)\bigg][/tex]
[tex]\text{Unit Circle:}\quad 2(0)\bigg[ \cos \bigg(\dfrac{7\pi}{26}\bigg)+ \cos \bigg(\dfrac{3\pi}{26}\bigg)\bigg][/tex]
Product of Zero 0
LHS = RHS: 0 = 0 [tex]\checkmark[/tex]
