By differentiating the identity cos(x+ alpha) = cos x cos

By differentiating the identity $\cos(x+\alpha) = \cos x\cos\alpha - \sin x\sin\alpha, \tag{1}$ show that $\sin(x+\alpha) = \sin x\cos\alpha + \cos x\sin\alpha$.

Step 1: Differentiate the left-hand side

Treat $\alpha$ as a fixed parameter and differentiate with respect to $x$. By the chain rule with $u = x+\alpha$, $u' = 1$: $\frac{d}{dx}\cos(x+\alpha) = -\sin(x+\alpha)$.

Step 2: Differentiate the right-hand side

Here $\cos\alpha$ and $\sin\alpha$ are constants with respect to $x$, so they simply ride along: \[\frac{d}{dx}\bigl(\cos x\cos\alpha - \sin x\sin\alpha\bigr) = -\sin x\cos\alpha - \cos x\sin\alpha.\]

Step 3: Equate and tidy

Identity (1) holds for every $x$, so the two derivatives are equal: $-\sin(x+\alpha) = -\sin x\cos\alpha - \cos x\sin\alpha$.

Multiplying through by $-1$: $\sin(x+\alpha) = \sin x\cos\alpha + \cos x\sin\alpha$.

Step 4: Verification

\[\alpha = \frac{\pi}{2}:\quad \sin\!\left(x+\tfrac{\pi}{2}\right) = \sin x\cdot 0 + \cos x\cdot 1 = \cos x \;\;\checkmark\]

\[x = \alpha = \frac{\pi}{6}:\quad \sin\frac{\pi}{3} = \frac{\sqrt3}{2}; \qquad \frac12\cdot\frac{\sqrt3}{2} + \frac{\sqrt3}{2}\cdot\frac12 = \frac{\sqrt3}{2} \;\;\checkmark\]