Question

Arata că: sin(3π/2+x)-sin(3π/2-x)=0
Va rog ajutati-ma

Answer 1

sinα=cos(π/2-α)
cosα=cos(-α)=cos(2π-α)
sinπ=0
atunci

cos (π/2-3π/2-x)-cos(π/2-3π/2+x)=cos(-π-x)-cos(-π+x) =cos(π+x)-cos(π-x)=
cosπcosx-sinπsinx-(cosπcosx+sinπsinx)=cosx-0-cosx-0=0

altfel, direct, mai rapid
tinand cont ca sin3π/2=-1
sin3π/2cosx+sinxcos3π/2-(sin3π/2cosx-sinxcos3π/2)=
-cosx+0-(-cosx-0)=-cosx+cosx=0

Answer 2

Cam asa se rezolva cel mai usor... folosind formula sina-sinb=2sin(a-b)/2 * cos(a+b)/2