product of sums: $R_{XX}(t_1, t_2) = cos(t_2 - t_2) + cos t_1 cos t_2$

starting with this:

$$R_{XX}(t_1, t_2) = 2 cos t_1 cos t_2 + sin t_1 sin t_2$$

The textbook say it should be reducible to this form:

$$R_{XX}(t_1, t_2) = cos(t_2 – t_2) + cos t_1 cos t_2$$

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So I try to do the reduction using the following identities:

$$cos A cos B = frac{1}{2}cos(A-B) + frac{1}{2}cos(A+B)$$

$$sin A sin B = frac{1}{2}cos(A-B) – frac{1}{2}cos(A+B)$$

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Here’s my work:

$$R_{XX}(t_1, t_2) = 2 cos t_1 cos t_2 + sin t_1 sin t_2$$

$$begin{aligned}R_{XX}(t_1, t_2) &= 2 (frac{1}{2}cos(t_1-t_2) + frac{1}{2}cos(t_1+t_2)) \ &+ frac{1}{2}cos(t_1 – t_2) – frac{1}{2}cos(t_1 + t_2)end{aligned}$$

$$begin{aligned}R_{XX}(t_1, t_2) &= cos(t_1-t_2) + cos(t_1+t_2)) \ &+ frac{1}{2}cos(t_1 – t_2) – frac{1}{2}cos(t_1 + t_2)end{aligned}$$

$$boxed{R_{XX}(t_1, t_2) = frac{3}{2}cos(t_1 – t_2) +frac{1}{2}cos(t_1+t_2)}$$

Any ideas why it doesn’t equal?

$$R_{XX}(t_1, t_2) = cos(t_2 – t_2) + cos t_1 cos t_2$$