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

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(D) Given,$\sin \left( x + \frac{4\pi}{9} \right) = a$. Note that $\frac{4\pi}{9} = 80^{\circ}$.We need to find $\cos \left( x + \frac{7\pi}{9} \right

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

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(D) Given,$\sin \left( x + \frac{4\pi}{9} \right) = a$. Note that $\frac{4\pi}{9} = 80^{\circ}$.We need to find $\cos \left( x + \frac{7\pi}{9} \right)$. Note that $\frac{7\pi}{9} = 140^{\circ}$.We can write $\cos \left( x + 140^{\circ} \right) = \cos \left( (x + 80^{\circ}) + 60^{\circ} \right)$.Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:$\cos \left( (x + 80^{\circ}) + 60^{\circ} \right) = \cos(x + 80^{\circ}) \cos 60^{\circ} - \sin(x + 80^{\circ}) \sin 60^{\circ}$.Since $\frac{\pi}{9} In this interval,$\cos(x + 80^{\circ})$ is negative,so $\cos(x + 80^{\circ}) = -\sqrt{1 - \sin^2(x + 80^{\circ})} = -\sqrt{1 - a^2}$.Substituting the values: $\cos(x + 140^{\circ}) = (-\sqrt{1 - a^2}) \cdot \frac{1}{2} - a \cdot \frac{\sqrt{3}}{2} = \frac{-\sqrt{1 - a^2} - a\sqrt{3}}{2}$.

If $\sin \left( x + \frac{4\pi}{9} \right) = a$ and $\frac{\pi}{9} < x < \frac{\pi}{3}$,then $\cos \left( x + \frac{7\pi}{9} \right)$ equals:

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