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:


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?


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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Allawonder Done. $\endgroup$ – N. F. Taussig Aug 29 at 19:39
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.