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If $\sin \left( {x + \frac{{4\pi }}{9}} \right) = a;\,$ $\frac{\pi }{9}\, < \,x\, < \,\frac{\pi }{3},$ then $\cos \left( {x + \frac{{7\pi }}{9}} \right)$ equals :-
$\frac{{\sqrt {(1 - {a^2})} - a\sqrt 3 }}{2}$
$\frac{{1 - {a^2} + a\sqrt 3 }}{2}$
$\frac{{a\sqrt 3 - \sqrt {(1 - {a^2})} }}{2}$
$\frac{{ - \sqrt {(1 - {a^2})} - a\sqrt 3 }}{2}$
Solution
Given that $\sin \left(x+80^{\circ}\right)=a$
$\therefore \cos \left(x+140^{\circ}\right)=\cos \left\{\left(x+80^{\circ}\right)+60^{\circ}\right\}$
${=\cos \left(x+80^{\circ}\right) \cos 60^{\circ}-\sin \left(x+80^{\circ}\right) \sin 60^{\circ}} $
${=-\sqrt{\left(1-a^{2}\right)} \cdot \frac{1}{2}-\frac{a \sqrt{3}}{2}=\frac{-\sqrt{\left(1-a^{2}\right)}-\sqrt{3} a}{2}}$
$\left(\therefore 20^{\circ}<\mathrm{x}<60^{\circ}, \cos \left(\mathrm{x}+80^{\circ}\right)\right.is -ve)$
