Question

considerando x diferente de y a expressão sen(x+y).sen(x-y)é equivalente a? me ajudeeem por favor!

Answer 1

$E=sen(x+y)*sen(x-y)$
$E=(senx*cosy+seny*cosx)*(senx*cosy-seny*cosx)$
$E=(senx*cosy)^2-(seny*cosx)^2$

$E=sen^2x*cos^2y-sen^2y*cos^2x$

Answer 2

(Ufsm 2002) Considerando x ≠ y, a expressão sen(x + y).sen(x - y) é equivalente a ?

$\begin{array}{lcl}\sin(x + y)\cdot\sin(x - y) &=& \sin(x + y)\cdot\sin(x - y) \\~\\&=& \left[\sin(x)\cos(y)+\sin(y)\cos(x)\right]\cdot\left[\sin(x)\cos(-y)+\sin(-y)\cos(x)\right]\\~\\&=&\left[\sin(x)\cos(y)+\sin(y)\cos(x)\right]\cdot\left[\sin(x)\cos(y)-\sin(y)\cos(x)\right]\\~\\&=&\sin^2(x)\cos^2(y)-\sin^2(y)\cos^2(x)\\~\\&=& \cos^2(y)\cdot[1-\cos^2(x)]-\sin^2(y)\cos^2(x)\\~\\&=& \cos^2(y)-\cos^2(y)\cos^2(x)-\sin^2(y)\cos^2(x) \\~\\&=& \cos^2(y)-\cos^2(x)\cdot[\cos^2(y)+\sin^2(y)]\end{array}$$\begin{array}{lcl}\sin(x + y)\cdot\sin(x - y) &=& \cos^2(y) -\cos^2(x)\end{array}$