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10. Prove the following:

a. [tex]\cos^2 \theta + \sin^2 \theta \cdot \cos 2\alpha = \cos^2 \alpha + \sin^2 \alpha \cdot \cos 2\theta[/tex]

Sagot :

GinnyAnswer
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.
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