# Simplifying $\sin(\frac{7A}{2}+15^{\circ})\sin(\frac{3A}{2}-75^{\circ})+\cos(\frac{7A}{2}+15^{\circ})\cos(\frac{3A}{2}-75^{\circ})$

Can someone help me with simplifying this expression:

$$\sin \left( \dfrac{7A}{2} + 15^{\circ} \right) \sin \left( \dfrac{3A}{2} - 75^{\circ} \right) + \cos \left( \dfrac{7A}{2} + 15^{\circ} \right) \cos \left( \dfrac{3A}{2} - 75^{\circ} \right)$$

I know that $$\sin x \sin y + \cos x \cos y = \cos(x-y)$$, but this way I only get the expression:

$$\cos(2A+100^{\circ})$$

The problem is that I have to find the correct solution among the following ones:

• $$-2\sin A \cos A$$

• $$\cos ^2 A - \sin ^2 A$$

• $$- \sin A$$

• $$\cos A$$

Maybe I shouldn't have used this trigonometric identity but another one?

If we set $$x = \frac{3A}{2} - 75^\circ$$ and $$y = \frac{7A}{2} + 15^\circ$$ then $$x - y = -2A - 90^\circ$$ so \begin{align*} \cos(x - y) & = \cos(-2A - 90^\circ)\\ & = \cos(2A + 90^\circ) \end{align*} Can you take it from here?
• You needn't expand $\cos(2A+π/2)$ again. Just note that $\cos(\phi+π/2)=-\sin\phi.$ – Allawonder Aug 29 at 19:36
• In that case you should just have stopped at $\cos(2A+π/2).$ – Allawonder Aug 29 at 19:38
• @AleksandraAsanin The identity $\cos(x-y)=\cos x\cos y+\sin x\sin y$ is symmetric in its variables, so whichever of the arguments you set to $x$ and $y$ does not matter. You should get the same result if you make no mistakes. You do not include your work, but perhaps you're unconsciously taking $15$ to be $25.$ Such things happen. :) – Allawonder Aug 29 at 19:56