Cos3A+cos2A=sin(3A/2)+sin(A/2)?

Hello,

cos(3A) + cos(2A) = sin[(3A)/2] + sin(A/2) (0 < A < π)

let's apply the sum-to-product formula cos α + cos β = 2cos[(α + β)/2] cos[(α - β)/2]:

2cos[(3A + 2A)/2] cos[(3A - 2A)/2] = sin[(3A)/2] + sin(A/2)

2cos[(5A)/2] cos(A/2) = sin[(3A)/2] + sin(A/2)

let's apply the sum-to-product formula sin α + sin β = 2sin[(α + β)/2] cos[(α - β)/2]:

2cos[(5A)/2] cos(A/2) = 2sin{[(3A)/2] + (A/2)} /2} cos{[(3A)/2] - (A/2)} /2}

2cos[(5A)/2] cos(A/2) = 2sin{[(4A)/2] /2} cos{[(2A)/2] /2}

2cos[(5A)/2] cos(A/2) = 2sin[(2A)/2] cos(A/2)

2cos[(5A)/2] cos(A/2) = 2sinA cos(A/2)

(moving 2sinA cos(A/2) to the left side)

2cos[(5A)/2] cos(A/2) - 2sinA cos(A/2) = 0

let's factor out 2cos(A/2):

2cos(A/2) {cos[(5A)/2] - sinA} = 0

let's apply the co-function identity sinA = cos[(π/2) - A]:

2cos(A/2) {cos[(5A)/2] - cos[(π/2) - A]} = 0

let's apply the sum-to-product formula cos α - cos β = - 2sin[(α + β)/2] sin[(α - β)/2]:

2cos(A/2) {- 2sin{ {[(5A)/2] + [(π/2) - A]} /2} sin{ {[(5A)/2] - [(π/2) - A]} /2} } = 0

2cos(A/2) {- 2sin{ {[(5A)/2] + (π/2) - A} /2} sin{ {[(5A)/2] - (π/2) + A} /2} } = 0

- 4cos(A/2) sin{ {[(5A - 2A)/2] + (π/2)} /2} sin{ {[(5A + 2A)/2] - (π/2)} /2} = 0

- 4cos(A/2) sin{ {[(3A)/2] + (π/2)} /2} sin{ {[(7A)/2] - (π/2)} /2} = 0

- 4cos(A/2) sin{[(3A)/2](1/2) + (π/2)(1/2)} sin{[(7A)/2](1/2) - (π/2)(1/2)} = 0

- 4cos(A/2) sin{[(3A)/4] + (π/4)} sin{[(7A)/4] - (π/4)} = 0

let's now equate each factor to zero to find out the solutions:

cos(A/2) = 0 → A/2 = π/2 → A = 2(π/2) = π (nonetheless the endpoint π is not included)

sin{[(3A)/4] + (π/4)} = 0 → [(3A)/4] + (π/4) = 0 → (3A)/4 = - π/4 → 3A =
- π → A = - π/3;
moreover:
[(3A)/4] + (π/4) = π → (3A)/4 = π - (π/4) → (3A)/4 = (3π)/4 → A = π (nonetheless the endpoint π is not included)

sin{[(7A)/4] - (π/4)} = 0 → [(7A)/4] - (π/4) = 0 → (7A)/4 = π/4 → 7A = π → A = π/7;
moreover:
[(7A)/4] - (π/4) = π → (7A)/4 = π + (π/4) → (7A)/4 = (5π)/4 → 7A = 5π → A = (5π)/7
[(7A)/4] - (π/4) = 2π → (7A)/4 = 2π + (π/4) → (7A)/4 = (8π + π)/4 → (7A)/4 = (9π)/4 → 7A = 9π → A = (9π)/7


in conclusion, the only solutions such that 0 < A < π are:

A = π/7
A = (5π)/7


I hope it's helpful

Let a = A/2 , then

cos6a + cos4a = sin3a + sina
2 cos5a cosa = 2 sin2a cosa
cosa (cos5a - sin2a) = 0
cosa = 0 (rejected) or cos5a - sin2a = 0
sin(π/2 - 5a) = sin2a
2nπ + π/2 - 5a = 2a or 2nπ + π/2 - 5a = π - 2a
a = (2n + 1/2)π/7 or a = (2n - 1/2)π/3
A = (4n+1)π/7 or A = (4n-1)π/3 (rejected)
A = π/7 , 5π/7 for n = 0 , 1.

Thanks all