Question

Prove that ${\cos }^{-1}\frac{3}{11}-{\sin }^{-1}\frac{3}{4}={\sin }^{-1}\frac{19}{44}$

Answer 1

$\sin [{\cos }^{-1}\frac{3}{11}-{\sin }^{-1}\frac{3}{4}]=\frac{19}{44}$
$\alpha ={\cos }^{-1}\frac{3}{11},\beta ={\sin }^{-1}\frac{3}{4}$
$\therefore$ $\cos \alpha =\frac{3}{11}$
$\therefore$ $\sin \alpha =\sqrt{1-{\cos }^2\alpha }$
$=\sqrt{1-\frac{9}{121}}=\frac{\sqrt{112}}{11}=\frac{4\sqrt{7}}{11}$
Now $\sin \beta =\frac{3}{4}$
$\cos \beta =\sqrt{1-{\sin }^2\beta }$
$=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
Now from equation (1) taking LHS $=\sin (\alpha -\beta )$
$\sin \alpha \cos \beta -\sin \beta \cos \alpha$
On putting values
$=\frac{4\sqrt{7}}{11}\times \frac{\sqrt{7}}{4}-\frac{3}{4}\times \frac{3}{11}=\frac{28}{44}-\frac{9}{44}=\frac{19}{44}$
LHS $=$ RHS