Product to sum cos(3θ)cos(5θ)?

You're right; it does not matter. However, you need to remember that cos(-2θ) = cos(2θ) ≠ -cos(2θ). If you don't believe me, look at a cosine graph, which reflects over the y-axis. Visually speaking, cos(whatever) and cos(-whatever) on a unit circle are still the same x-coordinate. as they will be reflected over the x-axis

I think the book is right:-

cos[8t] = cos(5t)cos(3t)-sin(5t)sin(3t)

cos[2t] = cos(5t)cos(3t)+sin(5t)sin(3t)

add equations

cos(8t)+cos(2t) = 2cos(5t)cos(3t)

cos(5t)cos(3t) = (1/2)[cos(8t)+cos(2t)]

After your update

Update: product to sum using the formulas so

cosxcosy=(1/2)[cos(x-y)+cos(x+y)

it should be

cos(3θ)cos(5θ)

(1/2)[cos(3θ-5θ)+cos(3θ+5θ)]

(1/2)[cos(-2θ)+cos(8θ)]

Here is the mistake

(1/2)[-cos(2θ)+cos(8θ)]

cos(-2θ) = cos(2θ) it does not = -cos(2θ)

Remember that

cos(-2theta)=cos(2theta) !!!!!

--->cos3t*cos5t

=(1/2)[cos(3t-5t)+cos(3t+5t)]

=(1/2)[cos(-2t)+cos8t]

=(1/2)[cos2t+cos8t]

can you see?

It shouldn't. Just remember that cos(-t) = cos(t)

cos(3x)cos(5x) =>

cos(4x - x) * cos(4x + x) =>

(cos(4x)cos(x) + sin(4x)sin(x)) * (cos(4x)cos(x) - sin(4x)sin(x)) =>

(cos(4x)^2 * cos(x)^2 - sin(4x)^2 * sin(x)^2) =>

(cos(4x)^2 * cos(x)^2 - (1 - cos(4x)^2) * (1 - cos(x)^2)) =>

(cos(4x)^2 * cos(x)^2 - (1 - cos(4x)^2 - cos(x)^2 + cos(4x)^2 * cos(x)^2)) =>

(cos(4x)^2 * cos(x)^2 - 1 + cos(4x)^2 + cos(x)^2 - cos(4x)^2 * cos(x)^2) =>

(cos(x)^2 + cos(4x)^2 - 1) =>

((1/2) * (1 + cos(2x)) + (1/2) * (1 + cos(8x)) - 1) =>

(1/2) * (cos(2x) + cos(8x)) + (1/2) * 2 - 1 =>

(1/2) * (cos(2x) + cos(8x)) + 1 - 1 =>

(1/2) * (cos(2x) + cos(8x))