d/dx [3 (60^{} - x^{}) - 4 ³(30^{} + x^{})] = {1cm}{0.15mm}

Correct Answer is : π60cos⁡(3x∘)π/60 (3x^{})60πcos(3x∘)

Correct Answer is : π60cos⁡(3x∘)π/60 (3x^{})60πcos(3x∘)

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GUJCET 2025 Math Paper

Question : 22 of 40

Marks: +1, -0

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

Solution:

Concept:

Use the complement identity and the triple angle identity:

3\sin A - 4\sin^3 A = \sin 3A

For degree angles, convert

x^\circ

to radians before differentiating.

Explanation:

Let

A = 60^\circ - x^\circ

Since

30^\circ + x^\circ = 90^\circ - A

, we have

\cos(30^\circ + x^\circ)=\sin A

So the expression becomes

3\sin A -4\sin^3 A=\sin(3A)

Thus it equals

\sin[3(60^\circ-x^\circ)]=\sin(180^\circ-3x^\circ)

Using

\sin(180^\circ-\theta)=\sin\theta

, this is

\sin(3x^\circ)

Convert degrees:

3x^\circ = 3\cdot \frac{\pi x}{180}=\frac{\pi x}{60}

f(x)=\sin\left(\frac{\pi x}{60}\right)

Differentiate:

f'(x)=\cos\left(\frac{\pi x}{60}\right)\cdot \frac{\pi}{60}

Hence

f'(x)=\frac{\pi}{60}\cos(3x^\circ)

Answer:

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

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