¹This can be seen by simple trignormatic transformtion as follows: Starting with my solution, we write (I will remove the \(1,000\) factor from all values for simplicity, it makes no difference), so we have the following. My solution is

\[ h_{1}\left ( t\right ) =\operatorname{sinc}\left ( 2\pi \ t\right ) \cos \left ( 10\pi \ t\right ) \]

and key solution is

\[ h_{2}\left ( t\right ) =3\operatorname{sinc}\left ( 12\pi \ t\right ) -2\operatorname{sinc}\left ( 8\pi \ t\right ) \]

I will not convert \(h_{2}\left ( t\right ) \) to \(h_{1}\left ( t\right ) \)

\[ h_{2}\left ( t\right ) =3\frac{\sin \left ( 12\pi \ t\right ) }{12\pi \ t}-2\frac{\sin \left ( 8\pi \ t\right ) }{8\pi \ t}=\frac{\sin \left ( 12\pi \ t\right ) }{4\pi \ t}-\frac{\sin \left ( 8\pi \ t\right ) }{4\pi \ t}=\frac{1}{4\pi \ t}\left ( \sin \left ( 12\pi \ t\right ) -\sin \left ( 8\pi \ t\right ) \right ) \]

Now use \(\sin A-\sin B=2\sin \frac{A-B}{2}\cos \frac{A+B}{2}\) on the above, we obtain

\begin{align*} h_{2}\left ( t\right ) & =\frac{1}{4\pi \ t}\left ( 2\sin \frac{4\pi t}{2}\cos \frac{20\pi t}{2}\right ) =\frac{1}{2\pi \ t}\left ( \sin 2\pi t\cos 10\pi t\right ) \\ & =\frac{\sin 2\pi t}{2\pi \ t}\cos 10\pi t=\operatorname{sinc}\left ( 2\pi \ t\right ) \cos \left ( 10\pi \ t\right ) \end{align*}

which is \(h_{1}(t)\)