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

Question

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?

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

Answer 1

HINT Use the trigonometric identity

$$\cos(t_1-t_2)=\cos t_1\cos t_2+\sin t_1\sin t_2$$