If sin left( {x + frac{{4pi }}{9}} right) = a | vedclass

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Question

Given that $\sin \left(x+80^{\circ}ight)=a$

$	herefore \cos \left(x+140^{\circ}ight)=\cos \left\{\left(x+80^{\circ}ight)+60^{\circ}ight\}$

Vedclass · Accepted Answer

$\frac{{ - \sqrt {(1 - {a^2})}  - a\sqrt 3 }}{2}$

Vedclass · Answer

Given that $\sin \left(x+80^{\circ}ight)=a$

$	herefore \cos \left(x+140^{\circ}ight)=\cos \left\{\left(x+80^{\circ}ight)+60^{\circ}ight\}$

${=\cos \left(x+80^{\circ}ight) \cos 60^{\circ}-\sin \left(x+80^{\circ}ight) \sin 60^{\circ}} $

${=-\sqrt{\left(1-a^{2}ight)} \cdot \frac{1}{2}-\frac{a \sqrt{3}}{2}=\frac{-\sqrt{\left(1-a^{2}ight)}-\sqrt{3} a}{2}}$

$\left(	herefore 20^{\circ}<\mathrm{x}<60^{\circ}, \cos \left(\mathrm{x}+80^{\circ}ight)ight.is -ve)$

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 :-

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