Expressing trigonometric sums as products

Question

I know all the identities. But how do I express $\sin A -\cos B$ as a product? The four identities don’t lend themselves to this. Thanks.

Answer 1

$\cos B$ is essentially $\sin(\pi/2-B)$. So the sum to product formula for $\sin x-\sin y$ applies here:

$$\sin A-\sin(\pi/2-B)=2\cos\left(\dfrac{A-B+\pi/2}{2}\right)\cos\left(\dfrac{A+B-\pi/2}{2}\right)$$

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Aliter:

You can also write $\sin A$ as $\cos(\pi/2-A)$. So the formula for $\cos x-\cos y$ can also be used:

$$\cos(\pi/2 -A)-\cos B=-2\sin\left(\dfrac{\pi/2-A+B}{2}\right)\sin\left(\dfrac{\pi/2-A-B}{2}\right)$$