Answer :
To prove the given equation:
[tex]\[ \cos^2 \theta + \sin^2 \theta \cdot \cos(2\alpha) = \cos^2 \alpha + \sin^2 \alpha \cdot \cos(2\theta) \][/tex]
we will simplify and compare both sides of the equation. Let’s proceed with the steps to show the equality.
### Step 1: Simplify the left-hand side (LHS)
The left-hand side of the equation is:
[tex]\[ \cos^2 \theta + \sin^2 \theta \cdot \cos(2\alpha) \][/tex]
Recall the double angle identity for cosine:
[tex]\[ \cos(2\alpha) = 2 \cos^2 \alpha - 1 \][/tex]
Substitute this into the LHS:
[tex]\[ \cos^2 \theta + \sin^2 \theta (2 \cos^2 \alpha - 1) \][/tex]
Distribute [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \cos^2 \theta + 2 \sin^2 \theta \cos^2 \alpha - \sin^2 \theta \][/tex]
Rearrange to group similar terms:
[tex]\[ \cos^2 \theta - \sin^2 \theta + 2 \sin^2 \theta \cos^2 \alpha \][/tex]
We have:
[tex]\[ \cos^2 \theta - \sin^2 \theta + 2 \sin^2 \theta \cos^2 \alpha = \cos^2 \theta + \sin^2 \theta (\cos(2\alpha)) \][/tex]
Using [tex]\( \cos^2 \theta + \sin^2 \theta = 1 \)[/tex], this term simplifies to:
[tex]\[ 1 + (-2 \sin^2 \alpha \sin^2 \theta + 1 - \sin^2 \theta) \][/tex]
Hence, the simplified form of the LHS is:
[tex]\[ -2 \sin^2 \alpha \sin^2 \theta + 1 \][/tex]
### Step 2: Simplify the right-hand side (RHS)
The right-hand side of the equation is:
[tex]\[ \cos^2 \alpha + \sin^2 \alpha \cdot \cos(2\theta) \][/tex]
Using the identity for [tex]\( \cos(2\theta) \)[/tex]:
[tex]\[ \cos(2\theta) = 2 \cos^2 \theta - 1 \][/tex]
Substitute this into the RHS:
[tex]\[ \cos^2 \alpha + \sin^2 \alpha (2 \cos^2 \theta - 1) \][/tex]
Distribute [tex]\(\sin^2 \alpha\)[/tex]:
[tex]\[ \cos^2 \alpha + 2 \sin^2 \alpha \cos^2 \theta - \sin^2 \alpha \][/tex]
Rearrange to group similar terms:
[tex]\[ \cos^2 \alpha - \sin^2 \alpha + 2 \sin^2 \alpha \cos^2 \theta \][/tex]
Using [tex]\( \cos^2 \theta + \sin^2 \theta = 1 \)[/tex], this term simplifies to:
[tex]\[ 1 + (-2 \sin^2 \theta \sin^2 \alpha + 1 - \sin^2 \alpha) \][/tex]
Hence, the simplified form of the RHS is:
[tex]\[ -2 \sin^2 \theta \sin^2 \alpha + 1 \][/tex]
### Conclusion
Both the left-hand side (LHS) and right-hand side (RHS) simplify to the same expression:
[tex]\[ -2 \sin^2 \alpha \sin^2 \theta + 1 \][/tex]
Therefore, we have shown that:
[tex]\[ \cos^2 \theta + \sin^2 \theta \cdot \cos(2\alpha) = \cos^2 \alpha + \sin^2 \alpha \cdot \cos(2\theta) \][/tex]
This completes the proof that the given equation holds true.
[tex]\[ \cos^2 \theta + \sin^2 \theta \cdot \cos(2\alpha) = \cos^2 \alpha + \sin^2 \alpha \cdot \cos(2\theta) \][/tex]
we will simplify and compare both sides of the equation. Let’s proceed with the steps to show the equality.
### Step 1: Simplify the left-hand side (LHS)
The left-hand side of the equation is:
[tex]\[ \cos^2 \theta + \sin^2 \theta \cdot \cos(2\alpha) \][/tex]
Recall the double angle identity for cosine:
[tex]\[ \cos(2\alpha) = 2 \cos^2 \alpha - 1 \][/tex]
Substitute this into the LHS:
[tex]\[ \cos^2 \theta + \sin^2 \theta (2 \cos^2 \alpha - 1) \][/tex]
Distribute [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \cos^2 \theta + 2 \sin^2 \theta \cos^2 \alpha - \sin^2 \theta \][/tex]
Rearrange to group similar terms:
[tex]\[ \cos^2 \theta - \sin^2 \theta + 2 \sin^2 \theta \cos^2 \alpha \][/tex]
We have:
[tex]\[ \cos^2 \theta - \sin^2 \theta + 2 \sin^2 \theta \cos^2 \alpha = \cos^2 \theta + \sin^2 \theta (\cos(2\alpha)) \][/tex]
Using [tex]\( \cos^2 \theta + \sin^2 \theta = 1 \)[/tex], this term simplifies to:
[tex]\[ 1 + (-2 \sin^2 \alpha \sin^2 \theta + 1 - \sin^2 \theta) \][/tex]
Hence, the simplified form of the LHS is:
[tex]\[ -2 \sin^2 \alpha \sin^2 \theta + 1 \][/tex]
### Step 2: Simplify the right-hand side (RHS)
The right-hand side of the equation is:
[tex]\[ \cos^2 \alpha + \sin^2 \alpha \cdot \cos(2\theta) \][/tex]
Using the identity for [tex]\( \cos(2\theta) \)[/tex]:
[tex]\[ \cos(2\theta) = 2 \cos^2 \theta - 1 \][/tex]
Substitute this into the RHS:
[tex]\[ \cos^2 \alpha + \sin^2 \alpha (2 \cos^2 \theta - 1) \][/tex]
Distribute [tex]\(\sin^2 \alpha\)[/tex]:
[tex]\[ \cos^2 \alpha + 2 \sin^2 \alpha \cos^2 \theta - \sin^2 \alpha \][/tex]
Rearrange to group similar terms:
[tex]\[ \cos^2 \alpha - \sin^2 \alpha + 2 \sin^2 \alpha \cos^2 \theta \][/tex]
Using [tex]\( \cos^2 \theta + \sin^2 \theta = 1 \)[/tex], this term simplifies to:
[tex]\[ 1 + (-2 \sin^2 \theta \sin^2 \alpha + 1 - \sin^2 \alpha) \][/tex]
Hence, the simplified form of the RHS is:
[tex]\[ -2 \sin^2 \theta \sin^2 \alpha + 1 \][/tex]
### Conclusion
Both the left-hand side (LHS) and right-hand side (RHS) simplify to the same expression:
[tex]\[ -2 \sin^2 \alpha \sin^2 \theta + 1 \][/tex]
Therefore, we have shown that:
[tex]\[ \cos^2 \theta + \sin^2 \theta \cdot \cos(2\alpha) = \cos^2 \alpha + \sin^2 \alpha \cdot \cos(2\theta) \][/tex]
This completes the proof that the given equation holds true.
