Proving the Sum to Product Identities (without using the product to sum identities) I got this project for my students from another teacher, and it did not come with a key.  There are four versions, each one is proving a different Sum to Product identity, but they can't use the product to sum identity.  The instructions say "Using reciprocal, Pythagorean, sum/difference, double and half-angle identities verify the equation" then they are given one of the identities.
We have figured out the sine ones, but we haven't figured out the cosine ones.  This is what I have so far on one of them.  Can anyone help?

 A: Let's correct your calculation.
\begin{align*}
-2&\sin\left(\frac{\alpha + \beta}{2}\right)\sin\left(\frac{\alpha - \beta}{2}\right)\\
& = -2\sin\left(\frac{\alpha}{2} + \frac{\beta}{2}\right)\sin\left(\frac{\alpha}{2} - \frac{\beta}{2}\right)\\ 
& = -2\left[\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) + \cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\right]\left[\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) - \cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\right]\\
& = -2\left[\sin^2\left(\frac{\alpha}{2}\right)\cos^2\left(\frac{\beta}{2}\right) - \sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\cos\left(\frac{\beta}{2}\right)\right.\\
& \qquad \left. + \sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\cos\left(\frac{\beta}{2}\right)\right.
\left. - \cos^2\left(\frac{\alpha}{2}\right)\sin^2\left(\frac{\beta}{2}\right)\right]\\
& = -2\left[\sin^2\left(\frac{\alpha}{2}\right)\cos^2\left(\frac{\beta}{2}\right) - \cos^2\left(\frac{\alpha}{2}\right)\sin^2\left(\frac{\beta}{2}\right)\right]\\
& = -2\left[\left(\frac{1 - \cos\alpha}{2}\right)\left(\frac{1 + \cos\beta}{2}\right) - \left(\frac{1 + \cos\alpha}{2}\right)\left(\frac{1 - \cos\beta}{2}\right)\right]\\
& = -2\left[\frac{1 + \cos\beta - \cos\alpha - \cos\alpha\cos\beta}{4} - \frac{1 + \cos\alpha - \cos\beta - \cos\alpha\cos\beta}{4}\right]\\
& = -2\left[\frac{2\cos\beta - 2\cos\alpha}{4}\right]\\
& = \cos\alpha - \cos\beta
\end{align*}
A: We have that
$$\cos (A+B)=\cos A\cos B - \sin A\sin B$$
$$\cos (A-B)=\cos A\cos B + \sin A\sin B$$
therefore
$$\cos (A+B)+\cos (A-B)=2\cos A\cos B $$
that is

$$\cos \theta_1 +\cos \theta_2=2\cos \left(\frac{\theta_1+\theta_2}{2}\right)\cos \left(\frac{\theta_1-\theta_2}{2}\right) $$

and by subtracting$$\cos (A-B)-\cos (A+B)=2\sin A\sin B $$
that is

$$\cos \theta_1 -\cos \theta_2=-2\sin \left(\frac{\theta_1+\theta_2}{2}\right)\sin \left(\frac{\theta_1-\theta_2}{2}\right) $$

In a similar way we can use
$$\sin (A+B)=\sin A\cos B + \cos A\sin B$$
$$\sin (A-B)=\sin A\cos B - \cos A\sin B$$
