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
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