Question

cos(n+1)π.cos(n-1)π-sin(n+1)π.sin(n-1)π​

Answer 1

Answer:

Cosπ.2n

Step-by-step explanation:

Cos(n+1)π.Cos(n-1)π - Sin(n+1)π.Sin(n-1)

In this expression we extract common π

= π{Cos(n+1).Cos(n-1) - Sin(n+1).Sin(n-1)}

We know that ,

CosA.CosB - Sin A.SinB = Cos(A+B) Using this formula☝️☝☝☝☝☝☝☝

= π.Cos(n+1+n-1)

= π.Cos(2n)

= Cosπ.2n