Question

Ma poate ajuta cineva va rog cu acest exercițiu?​

Answer 1

Răspuns:

a)

Explicație pas cu pas:

$(1-cosxcosy)^{2} -(sinxsiny)^{2} =\\= (1 -cosxcosy-sinxsiny)(1-cosxcosy+sinxsiny)=\\ =[1-(cosxcosy+sinxsiny)][1-(cosxcosy-sinxsiny)]=\\ =[1-cos(x-y)][1-cos(x+y)]=\\\\ =2sin\frac{x-y}{2} sin\frac{x-y}{2} \cdot2sin\frac{x+y}{2} sin\frac{x+y}{2} =\\ \\ =4(sin\frac{x+y}{2}sin\frac{x-y}{2}) ^{2} =\\ \\ =4\cdot(\frac{cos(\frac{x+y}{2}-\frac{x-y}{2})-cos(\frac{x+y}{2}+\frac{x-y}{2}) }{2} )^{2} =\\ \\ =(cosy-cosx)^{2} =\\\\ =(cosx-cosy)^{2}$