cleffavw8

2022-03-29

Proving $\mathrm{cos}\left({w}_{1}t\right)+\mathrm{cos}\left({w}_{2}t\right)=2\mathrm{cos}\left(\frac{1}{2}\left({w}_{1}+{w}_{2}\right)t\right)\mathrm{cos}\left(\frac{1}{2}\left({w}_{1}-{w}_{2}\right)t\right)$

Lana Hamilton

Remember that
$\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\left(\alpha \right)\mathrm{cos}\left(\beta \right)-\mathrm{sin}\left(\alpha \right)\mathrm{sin}\left(\beta \right)$
and
$\mathrm{cos}\left(\alpha -\beta \right)=\mathrm{cos}\left(\alpha \right)\mathrm{cos}\left(\beta \right)+\mathrm{sin}\left(\alpha \right)\mathrm{sin}\left(\beta \right)$
so you can add up both equalities to get
$\mathrm{cos}\left(\alpha +\beta \right)+\mathrm{cos}\left(\alpha -\beta \right)=2\mathrm{cos}\left(\alpha \right)\mathrm{cos}\left(\beta \right)$
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
$\alpha +\beta ={\omega }_{1}t$
$\alpha -\beta ={\omega }_{2}t$
Which gives you
$\alpha =\frac{1}{2}\left({\omega }_{1}t+{\omega }_{2}t\right)$
$\beta =\frac{1}{2}\left({\omega }_{1}t-{\omega }_{2}t\right)$
which is exactly what you wanted to prove

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