The image shows a set of trigonometry problems. I will solve problem number 15: $(cos\theta + cos\alpha)^2 + (sin\theta + sin\alpha)^2 = ?$

1. Problem Description

The image shows a set of trigonometry problems. I will solve problem number 15: $(cos\theta + cos\alpha)^2 + (sin\theta + sin\alpha)^2 = ?$

2. Solution Steps

First, we expand the squares:

$(cos\theta + cos\alpha)^2 + (sin\theta + sin\alpha)^2 = cos^2\theta + 2cos\theta cos\alpha + cos^2\alpha + sin^2\theta + 2sin\theta sin\alpha + sin^2\alpha$

Next, we group terms using the Pythagorean trigonometric identity $sin^2x + cos^2x = 1$:

$(cos^2\theta + sin^2\theta) + (cos^2\alpha + sin^2\alpha) + 2cos\theta cos\alpha + 2sin\theta sin\alpha = 1 + 1 + 2(cos\theta cos\alpha + sin\theta sin\alpha)$

Then, we apply the cosine subtraction identity $cos(A - B) = cosAcosB + sinAsinB$:

$2 + 2cos(\theta - \alpha) = 2[1 + cos(\theta - \alpha)]$

Using the half-angle identity $cos(2x) = 2cos^2(x) - 1$, we have $1 + cos(2x) = 2cos^2(x)$. Thus, $1 + cos(\theta - \alpha) = 2cos^2(\frac{\theta - \alpha}{2})$.

So, the expression becomes:

$2[2cos^2(\frac{\theta - \alpha}{2})] = 4cos^2(\frac{\theta - \alpha}{2})$

3. Final Answer

$4cos^2(\frac{\theta - \alpha}{2})$