The following questions may kindly be answered.

Asked by tps.mjmdr | 16th Jun, 2018, 08:37: PM

Expert Answer:

cosθ + cos(∏+θ) + .....+cos(n∏+θ)
Consider,
cos(∏+θ) = cos∏cosθ - sin∏sinθ
As sinn∏ is always zero and cos∏ = -1
cos(∏+θ) = -cosθ
similarly cos2∏ = 1
Now we have two cases 
i) n is odd
then cos(n∏+θ) = (-1)n cosθ = -cosθ
ii) n is even 
then cos(n∏+θ) = (-1)n cosθ = cosθ
USe this and solve it further

Answered by Sneha shidid | 19th Jun, 2018, 11:04: AM