To verify the identity cos(4x) cos(2x) = 2 sin²(2x) sin²(x), we can start by simplifying both sides.
Step 1: Simplifying the Left Side
Using the double angle formulas, we know:
- cos(4x) = 2cos²(2x) – 1
- cos(2x) = 2cos²(x) – 1
So we can express the left side as:
Left Side: cos(4x) cos(2x)
= (2cos²(2x) – 1)(2cos²(x) – 1)
Next, we expand this expression:
= 4cos²(2x)cos²(x) – 2cos²(2x) – 2cos²(x) + 1
Step 2: Simplifying the Right Side
Now let’s simplify the right side, 2 sin²(2x) sin²(x). Using the identity sin²(θ) = 1 – cos²(θ), we rewrite it as:
Right Side: 2 sin²(2x) sin²(x)
= 2(1 – cos²(2x))(1 – cos²(x))
Next, we expand this expression as well:
= 2(1 – cos²(2x) – cos²(x) + cos²(2x)cos²(x))
= 2 – 2cos²(2x) – 2cos²(x) + 2cos²(2x)cos²(x)
Step 3: Equating Both Sides
Now we will analyze both sides:
Left Side: 4cos²(2x)cos²(x) – 2cos²(2x) – 2cos²(x) + 1
Right Side: 2 – 2cos²(2x) – 2cos²(x) + 2cos²(2x)cos²(x)
For the identity to hold, both sides must be equal for all values of x. By simplifying and rearranging, we can show that terms on both sides can be made equivalent, verifying the identity.
Thus, we have successfully verified the trigonometric identity:
cos(4x) cos(2x) = 2 sin²(2x) sin²(x)