Trigonometry
posted by Kewal .
If cospΘ + cosqΘ = o. prove that the different values of Θ form two arithmetical progressions in which the common differences are 2π/p+q and 2π/pq respectively.

cos px = cos qx
cos px = cos ( pi  qx )
px = n ( pi ) + /  qx
taking the first case
px = n ( pi ) + qx
x ( p  q ) = n ( pi )
x = n ( ? p i) / ( p  q )
taking the second case we get
x = n ( ? pi ) / ( p + q )
We find the common difference is what is asked.