Evaluate: frac{d}{d x}left(3 cos left(frac{pi | vedclass

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(C) We know the trigonometric identity $\cos(3	heta) = 4\cos^3(	heta) - 3\cos(	heta)$.Given expression is $E = 3 \cos \left(\frac{\pi}{6}+x^{\circ}

Vedclass · Accepted Answer

$\frac{\pi}{60} \cos \left(3 x^{\circ}\right)$

Vedclass · Answer

(C) We know the trigonometric identity $\cos(3	heta) = 4\cos^3(	heta) - 3\cos(	heta)$.Given expression is $E = 3 \cos \left(\frac{\pi}{6}+x^{\circ}ight)-4 \cos ^3\left(\frac{\pi}{6}+x^{\circ}ight)$.This can be rewritten as $E = - (4 \cos ^3\left(\frac{\pi}{6}+x^{\circ}ight) - 3 \cos \left(\frac{\pi}{6}+x^{\circ}ight))$.Using the identity,$E = - \cos \left(3 \left(\frac{\pi}{6}+x^{\circ}ight)ight) = - \cos \left(\frac{\pi}{2} + 3x^{\circ}ight)$.Since $\cos(\frac{\pi}{2} + 	heta) = -\sin(	heta)$,we have $E = -(-\sin(3x^{\circ})) = \sin(3x^{\circ})$.Now,we need to differentiate $E$ with respect to $x$.First,convert $x^{\circ}$ to radians: $x^{\circ} = x 	imes \frac{\pi}{180}$ radians.So,$E = \sin\left(3 	imes \frac{\pi x}{180}ight) = \sin\left(\frac{\pi x}{60}ight)$.Now,$\frac{d}{dx} \sin\left(\frac{\pi x}{60}ight) = \cos\left(\frac{\pi x}{60}ight) 	imes \frac{d}{dx} \left(\frac{\pi x}{60}ight) = \frac{\pi}{60} \cos\left(\frac{\pi x}{60}ight)$.Converting back to degrees,$\frac{\pi x}{60} = 3x^{\circ}$.Thus,the derivative is $\frac{\pi}{60} \cos(3x^{\circ})$.Therefore,the correct option is $C$.

Evaluate: $\frac{d}{d x}\left(3 \cos \left(\frac{\pi}{6}+x^{\circ}\right)-4 \cos ^3\left(\frac{\pi}{6}+x^{\circ}\right)\right) = $ . . . . . .

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