Sabendo-se que, y1 = A cos(ωt)y2 = A cos(ωt + π/2)y3 ...

y3 = A√2 cos(ωt + 3π/4)

Sendo que:

Cos(a ± b) = cos a * cos b ± sin a * sin b

Aplicando em y3:

y3 = A√2 [cos(ωt ) * cos(3π/4) - sin(ωt) * sin(3π/4)]

Onde: cos(3π/4) = - √2 / 2 e sin(3π/4) = √2 / 2

y3 = A√2 [ - √2 / 2 * cos(ωt ) - sin(ωt) * √2 / 2 ]

y3 = - A cos(ωt ) - A sin(ωt)

y1 + y2 + y3 = A cos(ωt ) + A cos(ωt + π/2) - A cos(ωt ) - A sin(ωt)

y1 + y2 + y3 = A cos(ωt + π/2) - A sin(ωt)

e: -/+ sin(ωt) = cos(ωt +/- π/2)

y1 + y2 + y3 = A cos(ωt + π/2) + A cos(ωt + π/2)

y1 + y2 + y3 = 2 A cos(ωt + π/2)