Expand shifted angles before making small-angle approximations

Both copies give the full addition formula and warning. Small x does not make $x+\pi/6$ small, so replacing the whole sine by $x+\pi/6$ is inappropriate. Instead,

$\begin{aligned}&\sin(x+\pi/6)\\ & =\sin x\cos(\pi/6)+\cos x\sin(\pi/6).\end{aligned}$

The source special-angle values are $\cos(\pi/6)=\sqrt3/2$ and $\sin(\pi/6)=1/2$.

Use radians and $\sin x\approx x$, $\cos x\approx1-x^2/2$:

$\begin{aligned}&\sin(x+\pi/6)\\ & =\frac{\sqrt3}{2}\sin x+\frac12\cos x\approx\frac{\sqrt3}{2}x\\ &{}+\frac12\left(1-\frac{x^2}{2}\right)\\ & =\frac14(2+2\sqrt3x-x^2).\end{aligned}$

This is the complete printed approximation.

The source supplies both $\cos(A+B)=\cos A\cos B-\sin A\sin B$ and $\cos(A-B)=\cos A\cos B+\sin A\sin B$. Thus

$\cos(\pi/3+x)-\cos(\pi/3-x)$

$=[\cos(\pi/3)\cos x-\sin(\pi/3)\sin x]-[\cos(\pi/3)\cos x+\sin(\pi/3)\sin x]$

$\begin{aligned}&=-2\sin(\pi/3)\sin x\approx-2\left(\frac{\sqrt3}{2}\right)x\\ & =-\sqrt3x.\end{aligned}$

Both copies stress the distinction between = for an exact identity and ≈ for an approximation. Keep exact angle identities before introducing the small-angle truncations.