Prove that
sin(n+1)Asin(n-1)A + cos(n+1)Acos(n-1)A = cos2A
Let (n+1)A = x
Let (n-1)A = y
sinx*siny + cosx*cosy = cos (x - y)
(This is an identity you should know)
Now note that x -y = 2A , so
sinx*siny + cosx*cosy = cos (x - y)
= cos(2A)
To prove the given equality, we can use the trigonometric identity known as the angle addition formula:
cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
Let's use this identity to evaluate each side of the equation individually:
Left-hand side (LHS):
sin(n + 1)Asin(n - 1)A + cos(n + 1)Acos(n - 1)A
Using the angle addition formula, we can rewrite the above expression as:
sin(n + 1)Acos(n - 1)A + cos(n + 1)Asin(n - 1)A
Now, let's take a look at the right-hand side (RHS):
cos^2(A)
cos^2(A) can be expressed as:
cos(A)cos(A) - sin(A)sin(A)
Using the angle addition formula, we can rewrite the above expression as:
cos(A + A)
Comparing the LHS and RHS, we can see that they have the same form. By replacing (n + 1)A with A, and (n - 1)A with A, we get:
LHS = sin(A)cos(A) + cos(A)sin(A) = cos(A + A) = cos^2(A)
Therefore, we have proved that:
sin(n + 1)Asin(n - 1)A + cos(n + 1)Acos(n - 1)A = cos^2(A)