Express as a sum or difference. (Use the Product-to-Sum formulas.)

Question

Express as a sum or difference. (Use the Product-to-Sum formulas.)

(a). sin(−2x) cos 4x

(b). 3 cos x sin 7x

(c). cos 𝜃 − cos 9𝜃

Answer 1

For product to sum formulas we use identities

sin(u±v) = sin u cos v ± cos u sin v

cos(u±v) = cos u cos v (- or +) sin u sin v (there is no (- or +) symbol here, but it's opposite the ±)

So basically each identity has 2 separate equations

sin(u + v) = sin u cos v + cos u sin v

sin(u - v) = sin u cos v - cos u sin v

Adding both of those we come up with

sin(u + v) + sin(u - v) = 2 sin u cos v

which equates to

sin u cos v = 1/2[sin(u + v) + sin(u - v)]

So for part (a)

sin(-2x)cos(4x) = 1/2[sin(-2x + 4x) + sin(-2x - 4x)] = 1/2[sin(2x) + sin(-6x)]

Part (b)

3 cos x sin 7x = 3[sin 7x cos x] (u = 7x, v = x)

= 3(1/2[sin(7x + x) + sin(7x - x)] = 3/2[sin(8x) + sin(6x)]

Answer 2

This just involves a little bit of algebraic manipulation of the Sum to Product and Product to Sum Formulas.

a) Looking up the Product to Sum Formulas, it says that sin(u)cos(v) = 1/2[sin(u+v) + sin(u-v)]

If we make u = -2x and v = 4x, we have what we're looking for. From the above identity, we get

sin(-2x)cos(4x) = 1/2[sin(-2x+4x) + sin(-2x-4x)] = 1/2(sin(2x) + sin(-6x))

b) Consulting the same table, it says that cos(u)sin(v) = 1/2[sin(u+v) - sin(u-v)]

This means 3cos(u)sin(v) = 3/2[sin(u+v) - sin(u-v)]

If we let u = x and v = 7x, we have what we're looking for. Again looking at the above identity, we get that:

3cos(x)sin(7x) = 3/2[sin(x+7x) - sin(x-7x)] = 3/2(sin(8x) - sin(-6x))

c) Using another table of Sum to Product formulas, I get that cos(a) - cos(b) = -2sin((a+b)/2)sin((a-b)/2).

If we let a = θ and b = 9θ, we have what we're looking for. Using the above identity, we get that:

cos(θ) - cos(9θ) = -2sin((θ+9θ)/2)sin((θ-9θ)/2) = -2sin(10θ/2)sin(-8θ/2) = -2sin(5θ)sin(-4θ).