What is the answer of cos(n+1)a.cos(n-1)a+sin(n+1

The expression you provided can be simplified using trigonometric identities. Let's break it down step by step.

Understanding the Components

We have two parts in the expression:

• cos(n+1)a * cos(n-1)a
• sin(n+1)a * sin(n-1)a

Using the Cosine Addition Formula

We can apply the cosine addition formula:

cos(A)cos(B) = 0.5[cos(A+B) + cos(A-B)]

Applying this to our expression:

• Let A = (n+1)a and B = (n-1)a
• Then, cos(n+1)a * cos(n-1)a = 0.5[cos(2na) + cos(2a)]

Using the Sine Addition Formula

For the sine part, we use the sine addition formula:

sin(A)sin(B) = 0.5[cos(A-B) - cos(A+B)]

Thus:

• sin(n+1)a * sin(n-1)a = 0.5[cos(2a) - cos(2na)]

Combining the Results

Now, we can combine both results:

0.5[cos(2na) + cos(2a)] + 0.5[cos(2a) - cos(2na)]

Simplifying Further

This simplifies to:

• 0.5[2cos(2a)] = cos(2a)

The Final Answer

Therefore, the simplified answer to your expression is:

cos(2a)