2014/029) Given

2cos p+6cos q+7cos r+9cos s=0 and

2sin p−6sin q+7sin r−9sin s=0

Prove that 3cos(p+s)=7cos(q+r)

Solution

we have

2cos p+9cos s=−6cos q−7cos r

square and get


4cos^2 p+81cos^2 q + 36 cos p cos s=36cos^2 q +49cos^2 r −84cos q cos r...(1)

from 2nd relation given

2sin p−9sin s=6sin q−7sin r

square and get

4sin^2 p+81sin ^2 s - 36 sin p sin s=36sin ^2 q +49 sin ^2 r −84sin q sin r...(2)

and (1) and (2) to get
4(cos^2 p + sin^2 p) +81(cos^2 s + sin ^2 s) + 36 (cos p cos q- sin p sin s)=36(cos^2 q + sin ^2 q )+49 ( cos^2 r + sin ^2 r) + 84s( cos q cos r - sin q sin r)
or
4+81+36cos(p+s)=36+49+84cos(q+r)

or
36cos(p+s)=84cos(q+r)

or 3cos(p+s)=7cos(q+r)