Question

If $\alpha$ and $\beta$ are the solutions of $a\cos \theta +b\sin \theta =c$, then show that
$\sin \alpha +\sin \beta =2bc/\sin 2bc(a^2+b^2)$
$\sin \alpha \sin \beta =(c^2+a^2)/(a^2+b^2)$

Answer 1

From the given relation, we have
$a\cos \theta =c-b\sin \theta$
Square and change in terms of $\sin \theta$
$a^2(1-{\sin }^2\theta )=c^2-2bc\sin \theta +b^2{\sin }^2\theta$
$\therefore (a^2+b^2){\sin }^2\theta -2bc\sin \theta +(c^2-a^2)=0$
Its roots are $\sin \alpha$ and $\sin \beta$ as $\alpha$ and $\beta$ are the values of $\theta$ as given
$\therefore \sin \alpha +\sin \beta =\frac{2bc}{a^2+b^2}$ (Sum of roots)
$\sin \alpha \sin \beta =\frac{c^2-a^2}{a^2+b^2}$ (product of roots)