Question

sen(x + y) + sen(x -y)

Answer 1

Fórmula de prostaférese (transformação de soma em produto):

$\boxed{\begin{array}{c}\mathrm{sen\,}p+\mathrm{sen\,}q=2\,\mathrm{sen}\!\left(\dfrac{p+q}{2} \right )\!\cos\!\left(\dfrac{p-q}{2} \right ) \end{array}}$


Usando a fórmula acima, com

$p=x+y~~\text{ e }~~q=x-y$

temos,

$\mathrm{sen}(x+y)+\mathrm{sen}(x-y)=2\,\mathrm{sen}\!\left(\dfrac{(x+y)+(x-y)}{2} \right )\!\cos\!\left(\dfrac{(x+y)-(x-y)}{2} \right )\\\\\\ =2\,\mathrm{sen}\!\left(\dfrac{x+\diagup\!\!\!\! y+x-\diagup\!\!\!\! y}{2} \right )\!\cos\!\left(\dfrac{\diagup\!\!\!\! x+y-\diagup\!\!\!\! x+y}{2} \right )\\\\\\ =2\,\mathrm{sen}\!\left(\dfrac{\diagup\!\!\!\! 2x}{\diagup\!\!\!\! 2}\right)\cos\left(\dfrac{\diagup\!\!\!\! 2y}{\diagup\!\!\!\! 2} \right)\\\\\\ =\boxed{\begin{array}{c}2\,\mathrm{sen\,}x\cos y \end{array}}$


Bons estudos! :-)