# $2 \cos \alpha \cos \beta = \cos( \alpha + \beta) + \cos(\alpha −\beta)$

Here's my problem could someone show me what the steps are?

$$2 \cos \alpha \cos \beta = \cos( \alpha + \beta) + \cos( \alpha − \beta)$$

Here's my problem:

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To see something along the lines of the exercise, you need to know that, $$2\cos A\cos B=\cos (A+B)+\cos(A-B)$$

From here, now substitute, $A=\alpha$ and $B=\beta$ to see the result!

If you know that, $$\cos A+\cos B=2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)$$

Now set $A=\alpha+\beta$ and $B=\alpha-\beta$. Can you evaluate what $A+B$ and $A-B$ are to see your result?

Do you know that \begin{align}\cos(\alpha+\beta)&=\cos\alpha\cos\beta-\sin\alpha\sin\beta\tag{1}\\\cos(\alpha-\beta)&=\cos\alpha\cos \beta+\sin\alpha\sin\beta\tag{2}\end{align}

Add $(1)$ and $(2)$ to see your result.

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