Question

Transforme em produto
a) y = sen (a + b + c) - sen (a - b + c)

Answer 1

Identidade utilizada:

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


Então, utilizando a identidade acima, onde

$p=a+b+c,\;\;q=a-b+c$

temos


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