frac{d}{dx} [3 sin(60^{circ} - x^{circ}) - 4 | vedclass

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Question

(C) First,convert the degree measure to radians: $x^{\circ} = \frac{\pi x}{180}$.We use the identity $\cos(30^{\circ} + x^{\circ}) = \sin(90^{\circ} -

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) First,convert the degree measure to radians: $x^{\circ} = \frac{\pi x}{180}$.We use the identity $\cos(30^{\circ} + x^{\circ}) = \sin(90^{\circ} - (30^{\circ} + x^{\circ})) = \sin(60^{\circ} - x^{\circ})$.Let $A = 60^{\circ} - x^{\circ}$. The expression becomes $3 \sin A - 4 \sin^3 A$.Using the trigonometric identity $3 \sin A - 4 \sin^3 A = \sin(3A)$,we get $\sin(3(60^{\circ} - x^{\circ})) = \sin(180^{\circ} - 3x^{\circ}) = \sin(3x^{\circ})$.Now,express this in terms of radians: $f(x) = \sin(3 \cdot \frac{\pi x}{180}) = \sin(\frac{\pi x}{60})$.Differentiating with respect to $x$ using the chain rule: $\frac{d}{dx} [\sin(\frac{\pi x}{60})] = \cos(\frac{\pi x}{60}) \cdot \frac{\pi}{60} = \frac{\pi}{60} \cos(3x^{\circ})$.

$\frac{d}{dx} [3 \sin(60^{\circ} - x^{\circ}) - 4 \cos^3(30^{\circ} + x^{\circ})] = \rule{1cm}{0.15mm}$

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