Question

If $a,b,c$ and $d$ are four numbers in the interval $\left[0 , \pi \right]$ such that $sina+7sinb=4\left(sin c + 2 sin d\right)$ and $cosa+7cosb=4\left(cos c + 2 cos d\right)$ , then the numerical value of $\frac{7 cos \left(b - c\right)}{cos \left(a - d\right)}$ is

Answer 1

It is given that
$sina+7sinb=4sinc+8sind$
$\Rightarrow sina-8sind=4sinc-7sinb....\left(i\right)$
Also given that
$cosa-8cosd=4cosc-7cosb\ldots \left(ii\right)$
Squaring and adding Eq. (i) and (ii), we get
$1+8^{2}-16\left(sin a sin d + cos a cos d\right)$
$=16+49-56\left(sin b sin c + cos b cos c\right)$
$\Rightarrow -16cos\left(a - d\right)=-56cos\left(b - c\right)$
$\Rightarrow \frac{cos \left(b - c\right)}{cos \left(a - d\right)}=\frac{2}{7}$
$\Rightarrow \frac{7 cos \left(b - c\right)}{cos \left(a - d\right)}=2$