Question

$sin(n$
$sin(n + 1)a.sin(n + 2)a + cos(n + 1)a.cos(n + 2)a = cos a$
​

Answer 1

Step-by-step explanation:

We know the formula

sinA.sinB + Cos A.Cos B = Cos (A-B)

sin(n + 1)a.sin(n + 2)a +

cos(n +1)a.cos(n + 2)a = cos a

L.H.S

sin(n + 1)a.sin(n + 2)a +

cos(n +1)a.cos(n + 2)a

Cos.(n+1 -n -2)

Cos(1)

= 0