Question

If $\alpha$ and $\beta$ are the $so{l}^n$ of $a\cos \theta +b\sin \theta =c,$ Then show that $cos(\alpha -\beta )=\frac{2c^2-(a^2+b^2)}{a^2+b^2}$

Answer 1

$a\cos \theta +b\sin \theta =c⟹b\sin \theta =c-a\cos \theta$

Squaring on both sides

$b^2{\sin }^2\theta =c^2-2ac\cos \theta +a^2{\cos }^2\theta$

$(a^2+b^2){\cos }^2\theta -2ac\cos \theta +(c^2-b^2)=0$

$\cos \alpha \cos \beta =\frac{c^2-b^2}{a^2+b^2}$

$a\cos \theta +b\sin \theta =c⟹a\cos \theta =c-b\sin \theta$

Squaring on both sides

$a^2{\cos }^2\theta =c^2-2bc\sin \theta +b^2{\sin }^2\theta$

$(a^2+b^2){\sin }^2\theta -2bc\sin \theta +(c^2-a^2)=0$

$\sin \alpha \sin \beta =\frac{c^2-a^2}{a^2+b^2}$

$\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta =\frac{2c^2-(a^2+b^2)}{a^2+b^2}$