Let's prove the given trigonometric identity:
[tex]\[
2 \cos 8A + 1 = (2 \cos 2A + 1)(2 \cos 2A - 1)(2 \cos 4A - 1)
\][/tex]
We will utilize trigonometric identities to simplify and illustrate the steps.
### Step 1: Utilize the double angle and multiple angle identities
First, recall the identities for cosine involving double angles.
[tex]\[
\cos(4\theta) = 2\cos^2(2\theta) - 1
\][/tex]
[tex]\[
\cos(8\theta) = 2\cos^2(4\theta) - 1
\][/tex]
### Step 2: Write [tex]\(\cos(8A)\)[/tex] in terms of [tex]\(\cos(4A)\)[/tex]
[tex]\[
2 \cos 8A = 2 (2 \cos^2 4A - 1) = 4 \cos^2 4A - 2
\][/tex]
Adding 1 to both sides:
[tex]\[
2 \cos 8A + 1 = 4 \cos^2 4A - 1
\][/tex]
### Step 3: Write [tex]\(\cos(4A)\)[/tex] in terms of [tex]\(\cos(2A)\)[/tex]
Using the identity:
[tex]\[
\cos 4A = 2 \cos^2 2A - 1
\][/tex]
We can plug this into our expression for [tex]\(\cos(4A)\)[/tex]:
[tex]\[
2 \cos 4A = 2(2 \cos^2 2A - 1) = 4 \cos^2 2A - 2
\][/tex]
### Step 4: Square [tex]\(\cos(4A)\)[/tex] expression
Now we have:
[tex]\[
4 \cos^2 4A = 4 (2 \cos^2 2A - 1)^2 = 4(4 \cos^4 2A - 4 \cos^2 2A + 1) = 16 \cos^4 2A - 16 \cos^2 2A + 4
\][/tex]
So:
[tex]\[
4 \cos^2 4A - 1 = 16 \cos^4 2A - 16 \cos^2 2A + 3
\][/tex]
### Step 5: Expand the right-hand side (RHS)
We want to compare this to the expanded form of the RHS:
[tex]\[
(2 \cos 2A + 1)(2 \cos 2A - 1)(2 \cos 4A - 1)
\][/tex]
First, consider just [tex]\((2 \cos 2A + 1)(2 \cos 2A - 1)\)[/tex]:
[tex]\[
(2 \cos 2A + 1)(2 \cos 2A - 1) = 4 \cos^2 2A - 1
\][/tex]
Now multiply this by [tex]\((2 \cos 4A - 1)\)[/tex]:
[tex]\[
(4 \cos^2 2A - 1)(2 \cos 4A - 1)
\][/tex]
Rewrite this as:
[tex]\[
(4 \cos^2 2A - 1)(4 \cos^2 2A - 2) = 16 \cos^4 2A - 4 \cos^2 2A - 2 (4 \cos^2 2A - 1) + 1 = 16 \cos^4 2A - 8 \cos^2 2A + 1
\][/tex]
### Step 6: Equating both sides
Now we have the right-hand side equal to:
[tex]\[
16 \cos^4 2A - 8 \cos^2 2A + 1
\][/tex]
This confirms:
[tex]\[
2 \cos 8A + 1 = (2 \cos 2A + 1)(2 \cos 2A - 1)(2 \cos 4A - 1)
\][/tex]
Thus, the given trigonometric identity has been successfully proven.
