Answer: see proof below
Step-by-step explanation:
Use the following Sum Identities:
cos (A + B) = cosA · cosB - sinA · sinB
sin (A + B) = sinA · cos B - cosA · sinB
Use the Unit Circle to evaluate the following:
cos 30 = √3/2 sin 30 = 1/2
cos 45 = √2/2 sin 45 = √2/2
cos 120 = -1/2 sin 120 = √3/2
cos 240 = -1/2 sin 240 = -√3/2
cos 315 = √2/2 sin 315 = -√2/2
cos 330 = √3/2 sin 330 = -1/2
Proof LHS → RHS
[tex]\text{LHS:}\qquad \qquad \dfrac{\cos 285+\cos 345}{\sin 435-\sin 375}[/tex]
[tex]\text{Expand:}\qquad \quad \dfrac{\cos (240+45)+\cos (315+30)}{\sin (315+120)-\sin (330+45)}[/tex]
[tex]\text{Sum Identity:}\qquad \dfrac{\cos 240\cdot \cos 45-\sin 240\cdot 45+\cos 315\cdot \cos 30-\sin 315\cdot 30}{\sin 315\cdot \cos 120+\cos315\cdot \sin 120-(\sin330\cdot \cos45+\cos 330\cdot \sin 45) }[/tex]
[tex]\text{Unit Circle:}\quad \dfrac{(\frac{-1}{2}\cdot \frac{\sqrt2}{2})-(\frac{-\sqrt3}{2}\cdot \frac{\sqrt2}{2})+(\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{3})-(\frac{-\sqrt2}{2}\cdot \frac{1}{2})}{(\frac{-\sqrt2}{2}\cdot \frac{-1}{2})+(\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2)}-(\frac{-1}{2}\cdot \frac{\sqrt2}{2})-(\frac{\sqrt3}{2}\cdot \frac{\sqrt2}{2})}[/tex]
[tex]\text{Simplify:}\qquad \dfrac{-\sqrt2+\sqrt6+\sqrt6+\sqrt2}{\sqrt2+\sqrt6+\sqrt2-\sqrt6}\qquad =\dfrac{2\sqrt6}{2\sqrt2}\qquad =\sqrt3[/tex]
LHS = RHS: [tex]\sqrt3 = \sqrt3\qquad \checkmark[/tex]
