Question

(50 PONTOS) Simplifique a expressão:

$E=\cos\left(\dfrac{p-q}{2} \right )\cdot \left[A\cos\left(\dfrac{p+q}{2} \right )+B\,\mathrm{sen}\left(\dfrac{p+q}{2} \right ) \right ]+\\ \\ \\ -\mathrm{sen}\left(\dfrac{p-q}{2} \right )\cdot \left[A\,\mathrm{sen}\left(\dfrac{p+q}{2} \right )+B\cos\left(\dfrac{p+q}{2} \right ) \right ]$

Answer 1

Para simplificar a escrita, considere

$\dfrac{p+q}{2}=x,~\dfrac{p-q}{2}=y$
_______________________

$E=cos(y)\cdot[Acos(x)+Bsen(x)]-sen(y)\cdot[Asen(x)+Bcos(x)]\\\\E=Acos(x)cos(y)+Bsen(x)cos(y)-Asen(x)sen(y)-Bcos(x)sen(y)\\\\E=Acos(x)cos(y)-Asen(x)sen(y)+Bsen(x)cos(y)-Bsen(y)cos(x)\\\\E=A[cos(x)cos(y)-sen(x)sen(y)]+B[sen(x)cos(y)-sen(y)cos(x)]$

Note que temos as expansões do cosseno da soma e do seno da diferença, pois

$cos(x)cos(y)-sen(x)sen(y)=cos(x+y)=cos\left(\dfrac{p+q}{2}+\dfrac{p-q}{2}\right)\\\\\\\boxed{\boxed{cos(x)cos(y)-sen(x)sen(y)=cos(p)}}[$

e

$sen(x)cos(y)-sen(y)cos(x)=sen(x-y)=sen\left(\dfrac{p+q}{2}-\dfrac{p-q}{2}\right)\\\\\\\boxed{\boxed{sen(x)cos(y)-sen(y)cos(x)=sen(q)}}$

Então:

$\boxed{\boxed{E=Acos(p)+Bsen(q)}}$