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

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

Vedclass · Accepted Answer

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

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(D) Given $\sin \left( x + \frac{4\pi}{9} \right) = a$. We need to find $\cos \left( x + \frac{7\pi}{9} \right)$. Note that $\frac{7\pi}{9} = \frac{4\pi}{9} + \frac{3\pi}{9} = \frac{4\pi}{9} + \frac{\pi}{3}$. So,$\cos \left( x + \frac{7\pi}{9} \right) = \cos \left( \left( x + \frac{4\pi}{9} \right) + \frac{\pi}{3} \right)$. Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$: $\cos \left( \left( x + \frac{4\pi}{9} \right) + \frac{\pi}{3} \right) = \cos \left( x + \frac{4\pi}{9} \right) \cos \frac{\pi}{3} - \sin \left( x + \frac{4\pi}{9} \right) \sin \frac{\pi}{3}$. Given $\frac{\pi}{9} In this interval,the cosine function is negative. Since $\sin \left( x + \frac{4\pi}{9} \right) = a$,then $\cos \left( x + \frac{4\pi}{9} \right) = -\sqrt{1 - a^2}$. Substituting the values: $-\sqrt{1 - a^2} \cdot \frac{1}{2} - a \cdot \frac{\sqrt{3}}{2} = \frac{-\sqrt{1 - a^2} - a\sqrt{3}}{2}$.

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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