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Toolkit 40

Trigonometric Product-to-Sum Identities

sin⁡pcos⁡q=12(sin⁡(p+q)+sin⁡(p−q))\sin p\cos q=\frac{1}{2}(\sin(p+q)+\sin(p-q))
cos⁡pcos⁡q=12(cos⁡(p+q)+cos⁡(p−q))\cos p\cos q=\frac{1}{2}(\cos(p+q)+\cos(p-q))
sin⁡psin⁡q=12(cos⁡(p−q)−cos⁡(p+q))\sin p\sin q=\frac{1}{2}(\cos(p-q)-\cos(p+q))

Proof

By Toolkit 39,

sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2).\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right).

Let α=p+q\alpha = p+q and β=p−q\beta = p-q. Then

α+β2=p,α−β2=q.\frac{\alpha+\beta}{2} = p, \qquad \frac{\alpha-\beta}{2} = q.

Therefore,

sin⁡(p+q)+sin⁡(p−q)=2sin⁡pcos⁡q,\sin(p+q) + \sin(p-q) = 2\sin p\cos q,

so

sin⁡pcos⁡q=12(sin⁡(p+q)+sin⁡(p−q)).\sin p\cos q = \frac{1}{2}\left(\sin(p+q) + \sin(p-q)\right).

Similarly, by Toolkit 39,

cos⁡α+cos⁡β=2cos⁡(α+β2)cos⁡(α−β2).\cos\alpha + \cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right).

Using the same substitutions,

cos⁡(p+q)+cos⁡(p−q)=2cos⁡pcos⁡q.\cos(p+q) + \cos(p-q) = 2\cos p\cos q.

Hence,

cos⁡pcos⁡q=12(cos⁡(p+q)+cos⁡(p−q)).\cos p\cos q = \frac{1}{2}\left(\cos(p+q) + \cos(p-q)\right).

Finally, by Toolkit 39,

cos⁡α−cos⁡β=−2sin⁡(α+β2)sin⁡(α−β2).\cos\alpha - \cos\beta = -2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right).

Interchanging the order on the left and again using α=p+q\alpha = p+q, β=p−q\beta = p-q, gives

cos⁡(p−q)−cos⁡(p+q)=2sin⁡psin⁡q.\cos(p-q) - \cos(p+q) = 2\sin p\sin q.

Therefore,

sin⁡psin⁡q=12(cos⁡(p−q)−cos⁡(p+q)).□\sin p\sin q = \frac{1}{2}\left(\cos(p-q) - \cos(p+q)\right). \quad\square