# How do you use the product to sum formulas to write 5cos(-5beta)cos3beta as a sum or difference?

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#### Explanation

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#### Explanation:

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1
Feb 17, 2017

See below

#### Explanation:

Firstly, the sum of cosines is always written in the form $2 \cos \left(\frac{\text{P"+"Q")/2)cos(("P"-"Q}}{2}\right)$, so we have to take the $5$ out.

$5 \cos \left(- 5 \beta\right) \cos 3 \beta = \frac{5}{2} \left(2 \cos \left(- 5 \beta\right) \cos 3 \beta\right)$

Since $\cos x$ is an even function, we know that $\cos \left(- 5 \beta\right) = \cos \left(5 \beta\right)$

$\frac{5}{2} \left(2 \cos \left(- 5 \beta\right) \left(3 \beta\right)\right) = \frac{5}{2} \left(2 \cos 5 \beta \cos 3 \beta\right)$

Now it is trivial to find $\text{P}$ and $\text{Q}$

$\text{P} = 5 \beta + 3 \beta = 8 \beta$

$\text{Q} = 5 \beta - 3 \beta = 2 \beta$

$\frac{5}{2} \left(2 \cos 5 \beta \cos 3 \beta\right) = \frac{5}{2} \left(\cos 8 \beta + \cos 3 \beta\right)$

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