By differentiating the identity sin 3x = 3 cos²x sin x

By differentiating the identity $\sin 3x = 3\cos^{2}x\sin x - \sin^{3}x, \tag{1}$ show that $\cos 3x = \cos^{3}x - 3\cos x\sin^{2}x$.

Step 1: Differentiate the left-hand side

\[ \frac{d}{dx}\sin 3x = 3\cos 3x. \]

Step 2: Differentiate the first term on the right

Apply the product rule to $3\cos^{2}x\cdot\sin x$, with $\bigl(\cos^{2}x\bigr)' = -2\cos x\sin x$: \[\frac{d}{dx}\bigl(3\cos^{2}x\sin x\bigr) = 3\Bigl[-2\cos x\sin x\cdot\sin x + \cos^{2}x\cdot\cos x\Bigr] = 3\cos^{3}x - 6\cos x\sin^{2}x.\]

Step 3: Differentiate the second term

\[ \frac{d}{dx}\bigl(-\sin^{3}x\bigr) = -3\sin^{2}x\cos x. \]

Step 4: Combine and equate

Adding the two contributions, the $\cos x\sin^{2}x$ terms merge: $3\cos^{3}x - 6\cos x\sin^{2}x - 3\cos x\sin^{2}x = 3\cos^{3}x - 9\cos x\sin^{2}x$.

Equating with the left-hand side and dividing by $3$: $3\cos 3x = 3\cos^{3}x - 9\cos x\sin^{2}x \;\Longrightarrow\; \cos 3x = \cos^{3}x - 3\cos x\sin^{2}x$.

Step 5: Verification

Eliminating the sine with $\sin^{2}x = 1-\cos^{2}x$ recovers the familiar form: \[\cos^{3}x - 3\cos x\bigl(1-\cos^{2}x\bigr) = \cos^{3}x - 3\cos x + 3\cos^{3}x = 4\cos^{3}x - 3\cos x \;\;\checkmark\]

\[x = \frac{\pi}{3}:\quad \cos\pi = -1; \qquad 4\left(\tfrac12\right)^{3} - 3\cdot\tfrac12 = \tfrac12 - \tfrac32 = -1 \;\;\checkmark\]

The starting identity (1) is likewise the familiar $\sin 3x = 3\sin x - 4\sin^{3}x$ in disguise, since $3\cos^{2}x\sin x = 3\sin x - 3\sin^{3}x$.