Manipulating a trigonometric expression

Question

I'm sure this is simple, but I have not been able to see why the following expression is true:

$$\frac{-\cos{\theta}+\sqrt{3}\sin{\theta}}{2}=\cos {\left( \theta - \frac{2\pi}{3} \right) }$$

Thanks

Answer 1

From the left side:

$$\cos(\theta)\left(-\dfrac 12\right)+\sin(\theta)\left(\dfrac {\sqrt3}2\right)=\cos(\theta)\cos(\phi)+\sin(\theta)\sin(\phi)=\cos(\theta-\phi).$$

Can you find $\phi$ such that $\cos(\phi)=-\dfrac12$ and $\sin(\phi)=\dfrac {\sqrt3}2$?

Answer 2

Hint: $$\cos \left( \theta-\frac{2}{3}\pi \right) =-\cos \left( \theta+\frac{\pi}{3} \right)=-(\cos(\theta)\cos(\pi/3)-\sin(\theta)\sin(\pi/3))$$