Question

cos(n+1)alphacos(n-1)alpha+sin(n+1)alphasin(n-1)

Answer 1

Given: The expression cos(n+1)alphacos(n-1)alpha+sin(n+1)alphasin(n-1) alpha

To find: The value of the given expression.

Solution:

• Now we have given the expression as:

                  cos(n+1)alpha cos(n-1)alpha + sin(n+1)alpha sin(n-1) alpha

• Now we know the formula as:

                  cos a cos b + sin a sin b = cos (a - b)

• Here a = (n+1)alpha and b = (n-1)alpha

• So :

                  cos(a - b) = cos{ (n+1)alpha - (n-1)alpha }

                  cos(a - b) = cos{ (n(alpha) + alpha - n(alpha) + alpha }

                  cos(a - b) = cos{2 alpha}

Answer:

               So the value of given expression is cos{2 alpha}

Answer 2

Step-by-step explanation:

answer is cos 2alpha

thank you please mark me as brainlist