Let f be differentiable. Prove that the curves y=f(x)

Let $f$ be differentiable. Prove that the curves $y=f(x)$ and $y=f(x)\sin ax$ are tangent to each other at their common points.

Step 1: Common points

At a common point $x_0$: $f(x_0)=f(x_0)\sin ax_0$, i.e.\ $f(x_0)\bigl[1-\sin ax_0\bigr]=0$. This splits into two cases: either $\sin ax_0=1$, or $f(x_0)=0$ (possibly both).

Step 2: Slopes where $\sin ax_0=1$

Here $\cos ax_0=0$ as well, so

\[ \bigl(f(x)\sin ax\bigr)'\big|_{x_0} = f'(x_0)\underbrace{\sin ax_0}_{=1} + f(x_0)\cdot a\underbrace{\cos ax_0}_{=0} = f'(x_0) = \bigl(f(x)\bigr)'\big|_{x_0}. \tag{1} \]

So at every common point with $\sin ax_0=1$ — in particular this covers every common point with $f(x_0)\neq0$, since those are only possible via $\sin ax_0=1$ — the two curves are genuinely tangent.

Step 3: The other case: a zero of $f$ with $\sin ax_0\neq1$

If $f(x_0)=0$ but $\sin ax_0\neq1$, both curves still pass through $(x_0,0)$ — every curve of the form $y=f(x)\sin ax$ automatically vanishes wherever $f$ does, for any $a$. But there the slopes are

\[ \bigl(f(x)\bigr)'\big|_{x_0}=f'(x_0), \qquad \bigl(f(x)\sin ax\bigr)'\big|_{x_0} = f'(x_0)\sin ax_0, \tag{2} \] which agree only if $\sin ax_0=1$ (excluded by assumption here) or $f'(x_0)=0$ (i.e.\ $x_0$ is itself a double, not simple, zero of $f$). A curve like $f(x)=x$ with $a=1$ at $x_0=0$ makes this concrete: $f'(0)=1\neq0=f'(0)\sin0$, so $y=x$ and $y=x\sin x$ cross transversally at the origin, not tangentially.

Step 4: Verification

\[f(x)=\cos x,\ a=1,\ x_0=0\text{: } \sin0=0\neq1,\ f(0)=1\neq0 — \text{not a common point at all, consistent}\]

\[f(x)=x,\ a=1,\ x_0=0\text{: common point with } f(x_0)=0,\ \sin0=0\neq1;\ \text{slopes } 1 \text{ vs. } 0 \text{ — transversal, not tangent}\ \checkmark\]