Transforme em produto
a) y = sen (a + b + c) - sen (a - b + c)
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Então, utilizando a identidade acima, onde

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![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} } 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} }](https://tex.z-dn.net/?f=y%3D%5Cmathrm%7Bsen%7D%5Cleft%28a%2Bb%2Bc+%5Cright+%29-%5Cmathrm%7Bsen%7D%5Cleft%28a-b%2Bc+%5Cright+%29%5C%5C+%5C%5C+y%3D2%5Ccdot%5Cmathrm%7Bsen%7D%5Cleft%5B%5Cdfrac%7B%5Cleft%28a%2Bb%2Bc+%5Cright+%29-%5Cleft%28a-b%2Bc+%5Cright+%29%7D%7B2%7D+%5Cright+%5D%5Ccdot+%5Ccos%5Cleft%5B%5Cdfrac%7B%5Cleft%28a%2Bb%2Bc+%5Cright+%29%2B%5Cleft%28a-b%2Bc+%5Cright+%29%7D%7B2%7D+%5Cright+%5D%5C%5C+%5C%5C+y%3D2%5Ccdot%5Cmathrm%7Bsen%7D%5Cleft%5B%5Cdfrac%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+a%2Bb%2B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+c-%5Cdiagup%5C%21%5C%21%5C%21%5C%21+a%2Bb-%5Cdiagup%5C%21%5C%21%5C%21%5C%21+c%7D%7B2%7D+%5Cright+%5D%5Ccdot+%5Ccos%5Cleft%5B%5Cdfrac%7Ba%2B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+b%2Bc%2Ba-%5Cdiagup%5C%21%5C%21%5C%21%5C%21+b%2Bc%7D%7B2%7D+%5Cright+%5D%5C%5C+%5C%5C+y%3D2%5Ccdot+%5Cmathrm%7Bsen%7D%5Cleft%5B%5Cdfrac%7B2b%7D%7B2%7D+%5Cright+%5D%5Ccdot+%5Ccos%5Cleft%5B%5Cdfrac%7B2a%2B2c%7D%7B2%7D+%5Cright+%5D%5C%5C+%5C%5C+y%3D2%5Ccdot+%5Cmathrm%7Bsen%7D%5Cleft%5B%5Cdfrac%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+2%5Ccdot+b%7D%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+2%7D+%5Cright+%5D%5Ccdot+%5Ccos%5Cleft%5B%5Cdfrac%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+2%5Ccdot+%5Cleft%28a%2Bc+%5Cright+%29%7D%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21+2%7D+%5Cright+%5D%5C%5C+%5C%5C+%5C%5C+%5Cboxed%7B+%5Cbegin%7Barray%7D%7Bc%7D+y%3D2%5Ccdot+%5Cmathrm%7Bsen%5C%2C%7Db%5Ccdot+%5Ccos+%5Cleft%28a%2Bc+%5Cright%29+%5Cend%7Barray%7D+%7D)
Então, utilizando a identidade acima, onde
temos
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