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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?

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Your strategy is correct, but you made an arithmetic mistake.

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?

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  • $\begingroup$ You needn't expand $\cos(2A+π/2)$ again. Just note that $\cos(\phi+π/2)=-\sin\phi.$ $\endgroup$ – Allawonder Aug 29 at 19:36
  • $\begingroup$ @Allawonder Or I could let the OP figure that out. $\endgroup$ – N. F. Taussig Aug 29 at 19:37
  • $\begingroup$ In that case you should just have stopped at $\cos(2A+π/2).$ $\endgroup$ – Allawonder Aug 29 at 19:38
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    $\begingroup$ @Allawonder Done. $\endgroup$ – N. F. Taussig Aug 29 at 19:39
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    $\begingroup$ @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. :) $\endgroup$ – Allawonder Aug 29 at 19:56

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