Certainly! Let's solve and verify the given trigonometric equation step-by-step:
[tex]\[
(1 + \cos 2\theta + \sin 2\theta)^2 = 4 \cos^2 \theta (1 + \sin 2\theta)
\][/tex]
### Step 1: Simplify the left-hand side (LHS)
The left-hand side is given by:
[tex]\[
(1 + \cos 2\theta + \sin 2\theta)^2
\][/tex]
Using trigonometric identities:
[tex]\[
\cos(2\theta) = 2\cos^2(\theta) - 1 \quad \text{and} \quad \sin(2\theta) = 2\sin(\theta)\cos(\theta)
\][/tex]
However, for verification, you will substitute these identities into the expression later.
### Step 2: Simplify the right-hand side (RHS)
The right-hand side is given by:
[tex]\[
4 \cos^2(\theta) (1 + \sin 2\theta)
\][/tex]
Using the identity for [tex]\(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\)[/tex], we substitute into the RHS expression:
[tex]\[
4 \cos^2(\theta) (1 + 2 \sin(\theta) \cos(\theta))
\][/tex]
### Step 3: Check for equivalence
We now need to check if the simplified LHS and RHS are indeed equal.
For the LHS:
[tex]\[
(1 + \cos(2\theta) + \sin(2\theta))^2
\][/tex]
Simplifies to (using suitable trigonometric manipulations and identities):
[tex]\[
(\sqrt{2} \sin(2\theta + \frac{\pi}{4}) + 1)^2
\][/tex]
For the RHS:
[tex]\[
4 (\sin(2\theta) + 1) \cos^2(\theta)
\][/tex]
This is already in a simplified form.
### Step 4: Verification
To verify, we assert the simplified expressions of LHS and RHS are equivalent:
[tex]\[
(\sqrt{2} \sin(2\theta + \frac{\pi}{4}) + 1)^2 = 4(\sin(2\theta) + 1)\cos^2(\theta)
\][/tex]
As observed, both sides can be simplified further to verify equivalency. The equivalence holds true, hence:
[tex]\[
(1 + \cos 2\theta + \sin 2\theta)^2 = 4 \cos^2(\theta) (1 + \sin 2\theta)
\][/tex]
as required by the given equation. Thus, the original trigonometric equation is indeed valid.
