Can you prove this trigonometric equation? 3cos(p+s)=7cos(q+r)

[anemone]

Let $p,\,q,\,r,\,s\,\in[0,\,\pi]$ and we are given that

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

$2\sin p-6 \sin q+7 \sin r-9 \sin s=0$.

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

 

[kaliprasad]

Let $p,\,q,\,r,\,s\,\in[0,\,\pi]$ and we are given that

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

$2\sin p-6 \sin q+7 \sin r-9 \sin s=0$

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

Spoiler

we have

$2\cos p + 9\cos s = - 6 \cos q - 7 \cos r$

square and get
$4\cos^2 p + 81 \cos^2 s + 36 \cos p \cos s = 36 \cos ^2 q + 49 \cos ^2 r - 84 \cos q \cos r...(1)$
from 2nd relation given

$2\sin p-9 \sin s=6 \sin q- 7 \sin r$

we have sqaure and get

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

or
$4 + 81 + 36 \cos (p+ s) = 36 + 49 + 84\cos (q+r)$

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

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

 

[lfdahl]

Spoiler

\[2cos(p)+9cos(s)+6cos(q)+7cos(r)=0\\2sin(p)-9sin(s)-6sin(q)+7sin(r)=0\]

\[2cos(p)+9cos(s)=-6cos(q)-7cos(r)\\ 2sin(p)-9sin(s)=6sin(q)-7sin(r)\]

$$4cos^2(p)+81cos^2(s)+36cos(p)cos(s)=36cos^2(q)+49cos^2(r)+84cos(q)cos(r)\\ 4sin^2(p)+81sin^2(s)-36sin(p)sin(s)=36sin^2(q)+49sin^2(r)-84sin(q)sin(r)$$

Adding the two equations yields:
\[85 + 36cos(p+s)=85 + 84cos(q+r) \\

\Rightarrow 3cos(p+s) = 7cos(q+r)\]