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If cospθ+cosqθ=0, prove that the different values of θ form two arithmetical progressions in which the common difference are
2∏/(p+q) and 2∏/(p-q) resectively
we get
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(pie)
x=n(??pie)/p-q
taking the second case we get
x=n(??pie)/p+q;
we find the common difference is what is asked
note
pie denotes the constat 3.14....
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