Express as a product

Question

cos(a)+2cos(7a)+cos(13a)

I know the formulas but not how to apply them

Answer 1

The sum to product identity for cosine:

cos(x) + cos(y) = 2cos[(x+y)/2]cos[(x-y)/2]

The first and last terms of the expression in question are (middle term addressed later):

cos(a) + cos(13a)

Using the previously mentioned identity these two terms can be written as:

2cos(7a)cos(-6a), but since cos(-x) = cos(x) we have:

2cos(7a)cos(6a).

Now substitute that expression into the original:

2cos(7a)cos(6a) + 2cos(7a) (note the common factor 2cos(7a))

E = 2cos(7a)[cos(6a) + 1] (likely the intent is to express that last factor as one term)

We know cos(2x) = 2cos²(x) - 1 or:

2cos²(x) = cos(2x) + 1

That means for cos(6a) + 1 we can let 6a = 2x or x = 3a. That means:

2cos²(3a) = cos(6a) + 1

Back to the expression for E, replacing cos(6a) + 1:

E = 2cos(7a)[2cos²(3a)] = 4cos(7a)cos²(3a)

Visit this graph and use the slider on "a" to confirm:

desmos.com/calculator/krxeb6uhng