trigonometry - Proving $cos(w_1t) + cos(w_2t) = 2 cosleft(frac{1}{2}(w_1+w_2)tright) cos left(frac12 (w_1-w_2)tright)$

I am trying to prove :

$$\cos(w_1t) + \cos(w_2t) = 2 \cos\left(\frac{1}{2}(w_1+w_2)t\right) \cos \left(\frac12 (w_1-w_2)t\right)$$

My working so far uses the complex exponential identity:
$$\cos \theta = \frac12 \left(e^{i \theta} + e^{-i \theta}\right)$$

$$\begin{align} \frac12\left(e^{i w_1t} + e^{-i w_1t}\right) + \frac12\left(e^{i w_2t} + e^{-i w_2t}\right) &= \frac12\left(e^{it(w_1 + w_2)} + e^{-it(w_1 + w_2)}\right) \\ &= \cos\left((w_1 + w_2) t\right) \\ &= \cos\left(w_1t + w_2t\right) \end{align}$$

I cannot understand how to proceed further. Please help.

Answer

Remember that $$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$ and $$\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$$ so you can add up both equalities to get $$\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos(\alpha)\cos(\beta)$$ so now you want that $\alpha+\beta=\omega_{1}t$ and $\alpha-\beta=\omega_{2}t$ so you have to solve the system of equations $$\begin{array}{l} \alpha+\beta=\omega_{1}t\\ \alpha-\beta=\omega_{2}t \end{array}$$
Which gives you $$\begin{array}{l} \alpha=\dfrac{1}{2}\left(\omega_{1}t+\omega_{2}t\right)\\ \beta=\dfrac{1}{2}\left(\omega_{1}t-\omega_{2}t\right) \end{array}$$ which is exactly what you wanted to prove