If cos^3 80^{circ} + cos^3 40^{circ} - cos^3 | vedclass

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(B) We use the identity $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$,which implies $\cos^3 	heta = \frac{1}{4}(\cos 3	heta + 3\cos 	heta)$.Substi

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$\cos \left(\frac{2\pi}{3}\right)$

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(B) We use the identity $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$,which implies $\cos^3 	heta = \frac{1}{4}(\cos 3	heta + 3\cos 	heta)$.Substituting this into the expression $k = \cos^3 80^{\circ} + \cos^3 40^{\circ} - \cos^3 20^{\circ}$:$k = \frac{1}{4}(\cos 240^{\circ} + 3\cos 80^{\circ}) + \frac{1}{4}(\cos 120^{\circ} + 3\cos 40^{\circ}) - \frac{1}{4}(\cos 60^{\circ} + 3\cos 20^{\circ})$$k = \frac{1}{4} [(\cos 240^{\circ} + \cos 120^{\circ} - \cos 60^{\circ}) + 3(\cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ})]$Since $\cos 240^{\circ} = -\frac{1}{2}$,$\cos 120^{\circ} = -\frac{1}{2}$,and $\cos 60^{\circ} = \frac{1}{2}$,the first part is $(-\frac{1}{2} - \frac{1}{2} - \frac{1}{2}) = -\frac{3}{2}$.For the second part,$\cos 80^{\circ} + \cos 40^{\circ} = 2\cos 60^{\circ} \cos 20^{\circ} = 2(\frac{1}{2})\cos 20^{\circ} = \cos 20^{\circ}$.Thus,$3(\cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ}) = 3(\cos 20^{\circ} - \cos 20^{\circ}) = 0$.Therefore,$k = \frac{1}{4}(-\frac{3}{2} + 0) = -\frac{3}{8}$.Then $\frac{4k}{3} = \frac{4}{3} 	imes (-\frac{3}{8}) = -\frac{1}{2}$.Comparing with options: $\cos(\frac{2\pi}{3}) = \cos(120^{\circ}) = -\frac{1}{2}$.Hence,the correct option is $B$.

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

If $\cos^3 80^{\circ} + \cos^3 40^{\circ} - \cos^3 20^{\circ} = k$,then $\frac{4k}{3} =$

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