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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

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

Grade:12

2 Answers

HARIKRISHNAN M
18 Points
9 years ago

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....

Rajiv
15 Points
one year ago
Bro if cosx = cosy, then the general solution for all the angles included is 
x = 2npi +/- y.
Now according to your solution the question has not solved correctly.

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