Question

A expressão sen (a + b) . sen (a - b) é equivalente a:

a) cos b - cos a

b) sen b - sen a

c) cos²b - cos²a

d) sen²b - sen²a

e) sen (a²- b²)

Answer 1

$\begin{array}{c} \texttt{Vamos usar estas propriedades trigonom\'etricas:}\end{array}$


$\mathsf{\checkmark\quad sen~(a\pm b)=sen~a\cdot cos~b\pm sen~b\cdot cos~a)}$

$\mathsf{\checkmark\quad sen^{2}x +cos^{2}x=1}$

____________________


$\mathsf{sen~(a+b)\cdot sen~(a-b)}\\\\\\ \mathsf{=(sen~a\cdot cos~b+sen~b\cdot cos~a)\cdot (sen~a\cdot cos~b-sen~b\cdot cos~a)}\\\\\\ \mathsf{=(sen~a\cdot cos~b)^{2}-(sen~b\cdot cos~a)^{2}}\\\\\\ \mathsf{=sen^{2}~a\cdot cos^{2}~b-sen^{2}~b\cdot cos^{2}~a}$



$\begin{array}{c} \texttt{Substituindo} \mathtt{~(sen^{2}~a)} \texttt{~por} \mathtt{~(1-cos^{2}~a)}\end{array}$



$\mathsf{=(1-cos^{2}~a)\cdot cos^{2}~b-[(1-cos^{2}~b)\cdot cos^{2}~a]}\\\\\\ \mathsf{=cos^{2}~b-cos^{2}~a\cdot cos^{2}~b-[cos^{2}~a-cos^{2}~a\cdot cos^{2}~b]}\\\\\\ \mathsf{=cos^{2}~b-(cos^{2}~a\cdot cos^{2}~b)^{*}-cos^{2}~a+(cos^{2}~a\cdot cos^{2}~b)^{*}}\\\\\\ \mathsf{=cos^{2}~b-cos^{2}~a}\\\\\\ \boxed{\begin{array}{c} \mathsf{cos^{2}~b-cos^{2}~a}\end{array}}$


Alternativa: C


Bons estudos no Brainly! =)