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Can someone explain me please why

$2 \vec U \vec V cos(\omega t + \phi _1)cos(\omega t + \phi _2) = U V cos(\alpha )cos(2\omega t + \phi _1+\phi _2)+(cos\phi _1 - \phi _2) $,

where $\alpha$ is the angle between the vectors $\vec U$ and $\vec V$.

I just don't see it, It's from my physics book and I don't understand this step. I tried to use $cos(a+b) = cos(a)cos(b) - sin(a)sin(b)$ but I think it's nonsense because of that $sin$ and I didn't find any trigonometry identity which would help me neither.

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The identity does not hold true as posted, for example if $\omega = t = \phi_1 = \phi_2 = 0$ it reduces to $2 \vec U \vec V = UV \cos (\alpha)$ instead of the correct $2 \vec U \vec V = \color{red}{2}UV \cos (\alpha)\,$.

What does hold true, however, is: $$2 \vec U \vec V \cos(\omega t + \phi _1)\cos(\omega t + \phi _2) = U V \cos(\alpha )\big(\cos(2\omega t + \phi _1+\phi _2) \color{red}{+}\cos(\phi _1 - \phi _2)\big) $$

The above follows from the trigonometric identity $\cos(a+b)+\cos(a-b)=2 \cos(a)\cos(b)$.

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  • $\begingroup$ I have made a mistake, i will edit it and thank you very much! I have spent ages on this. $\endgroup$ – Leif Sep 11 '17 at 0:52

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