Answer: see proof below
Step-by-step explanation:
Use the following Double Angle identity: cos 2A = 2cos²A - 1
We are going to find one factor at a time:
2 cos16A + 1 = (2cos2A - 1)(2cos2A + 1)(2cos4A - 1)(2cos8A - 1)
↓ ↓ ↓ ↓
factor last factor 2nd factor 1st
Proof LHS → RHS:
Given: 2 cos16A + 1
let α = 8A, then 2·cos2α + 1
Identity: 2(2cos²α - 1) + 1
Distribute: 4cos²α - 2 + 1
Simplify: 4cos²α - 1
Factor: (2cosα - 1)(2cosα + 1)
Substitute: (2cos8A - 1)(2cos8A + 1)
let β = 4A, then 2·cos2β + 1
Identity: 2(2cos²β - 1) + 1
Distribute: 4cos²β - 2 + 1
Simplify: 4cos²β - 1
Factor: (2cosβ - 1)(2cosβ + 1)
Substitute: (2cos4A - 1)(2cos4A + 1)
let Ф = 4A, then 2·cos2Ф + 1
Identity: 2(2cos²Ф - 1) + 1
Distribute: 4cos²Ф - 2 + 1
Simplify: 4cos²Ф - 1
Factor: (2cosФ - 1)(2cosФ + 1)
Substitute: (2cos2A - 1)(2cos2A + 1)
(2cos2A - 1)(2cos2A + 1)(2cos4A - 1)(2cos8A - 1) = (2cos2A - 1)(2cos2A + 1)(2cos4A - 1)(2cos8A - 1) [tex]\checkmark[/tex]
