Prove that cos 80° – cos40° + $\sqrt{3}$ cos70° = 0
Solution:
cos 80° – cos40° + $\sqrt{3}$ cos70°
= 2 sin $\frac{80° + 40°}{2}$ sin $\frac{40° – 80°}{2}$ + $\sqrt{3}$ cos70°
= 2 sin $\frac{120}{2}$ sin $\left( – \frac{40}{2}\right)$ + $\sqrt{3}$ cos70°
= -2 sin 60° sin20°+ $\sqrt{3}$ cos70°
= -2 × $\frac{\sqrt{3}}{2}$ sin20° + $\sqrt{3}$ cos70°
= -$\sqrt{3}$ sin20° + $\sqrt{3}$ cos(90°-20°)
= -$\sqrt{3}$ sin20° + $\sqrt{3}$ sin 20°
= 0
∴ cos 80° – cos40° + $\sqrt{3}$ cos70° = 0 [Proved]