Question

Demostrar senB.cos(A-B)+cosB. sen(A-B)=senA

Answer 1

$Hola \\ \\ Demostrar~que:~~senB.cos(A-B)+cosB.sen(A-B)=senA \\ \\ \underline{Resoluci\acute{o}n~} \\ \\ senB[\underbrace{cosA.cosB+senA.senB}_{cos(A-B)}]+cosB.[\underbrace{senA.cosB-cosA.senB}_{sen(A-B)}]= \\ \\ senB.cosA.cosB+sen\²B.senA+senA.cos\²B-cosA.cosB.senB \\ \\ sen\²B.senA+senA. cos\²B~~~---\ \textgreater \ factorizando~[senA], tenemos: \\ \\ senA[\underbrace{sen\²A+cos\²A}_{1}]$

$\boxed{\boxed{senA}}~~---\ \textgreater \ queda~demostrado \\ \\ \mathbb{vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~haber~ayudado, saludos!!$