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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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) = $ . . . . . .

Similar Questions

$f(x)$ is differentiable on $\mathbb{R}$ and $f^{\prime}(m) \neq 0, \,m \in \mathbb{R}$. If $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$,then $m=$

If $h(x) = \sqrt{4f(x) + 3g(x)}$,$f(1) = 4$,$g(1) = 3$,$f'(1) = 4$,and $g'(1) = 3$,then find $h'(1)$.

If $f(x) = \frac{g(x) + g(-x)}{2} + \frac{2}{[h(x) + h(-x)]^{-1}}$,where $g$ and $h$ are differentiable functions,then find $f^{\prime}(0)$.

For the differentiable function $f: R - \{0\} \rightarrow R$,let $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$. Then $\left|f(3) + f^{\prime}\left(\frac{1}{4}\right)\right|$ is equal to

At $x=\frac{\pi^2}{4}$,find the value of $\frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module