Most of us have come across the impulse response of an ideal (continuous-time, unity gain) low pass filter with cut-off frequency $\omega_0$:
$$h_{LP}(t)=\frac{\sin(\omega_0t)}{\pi t}\tag{1}$$
The question is what happens if we replace the sine in the numerator of $(1)$ by a cosine? What type of filter is it? What is its frequency response?
$$\begin{align}h(t)&=\frac{\cos(\omega_0t)}{\pi t}\tag{2}\\H(\omega)&=\;?\end{align}$$
Please preface your answer with the spoiler tag >!.
I will use the Fourier Transform of the impulse response to determine the frequency response, noting that the impulse response given is non-causal with infinite time support. We can use the following Fourier Transform relationship (the impulse response for the Hilbert Transform): $$\frac{1}{\pi t} \leftrightarrow -j \text{sgn}(f)$$ Combined with the Fourier Transform of $\cos(\omega_o t) = \cos(2\pi f_o t)$: >!$$\cos(2\pi f_o t) \leftrightarrow \frac{1}{2}\delta(f-f_o) + \frac{1}{2}\delta(f+f_o)$$ The frequency response is then determined by convolving the two individual responses above (multiplication in time is convolution in frequency). Using superposition we can convolve each of the two impulses from the FT of the cosine with the FT of $1/(\pi t)$ (which I did in units of $f$ instead of $\omega$ to eliminate $2\pi$ terms using $\omega = 2\pi f$: $$H(f) = -j\text{sgn}(f) * \bigg(\frac{1}{2}\delta(f-f_o) + \frac{1}{2}\delta(f+f_o)\bigg) \\ = -j\text{sgn}(f) * \bigg(\frac{1}{2}\delta(f-f_o)\bigg) + j\text{sgn}(f) * \bigg(\frac{1}{2}\delta(f+f_o)\bigg) \\ = \frac{\text{sgn}(f-f_o)-\text{sgn}(f+f_o)}{2j} \\ =\begin{cases}-j,&f>f_0\\0,&-f_0<f<f_0\\j,&f<-f_0\end{cases}$$ Thus $$H(\omega) = \begin{cases}-j,&\omega>\omega_0\\0,&-\omega_0<\omega<\omega_0\\j,&\omega<-\omega_0\end{cases}$$This is depicted graphically below. The resulting frequency response has a magnitude that is the complement of the rectangular frequency response associated with a Sinc! Very interesting.
So this is a perfect "brickwall" High Pass Hilbert Transform filter! Note that this would also serve as a proof of the following identity, consistent with our intuition that a highpass can be formed by subtracting a low pass from a wire: $$\frac{\cos(\omega_o t)}{\pi t} = \mathscr{H}\bigg[\delta(t)-\frac{\sin(\omega_o t)}{\pi t}\bigg]$$
Dan's answer is correct and elegant. I would just like to explain another way of computing the Fourier transform of the given impulse response. In the summary further below I'll also mention a third simple way to arrive at the same result.
For the direct computation of the Fourier transform of the given impulse response we need the identity $$\int_{-\infty}^{\infty}\frac{\sin(\omega t)}{\pi t}dt=\textrm{sgn}(\omega)\tag{1}$$ Knowing that the integrand in $(1)$ is the impulse response of an ideal low pass filter with unity gain, it is clear that for positive $\omega$ the integral evaluates to $1$, because it equals the filter's DC gain. Negative $\omega$ just inverts the sign, and for $\omega=0$, the integrand vanishes. Hence, $(1)$ is established without any difficult math.
The Fourier transform of the given impulse response $h(t)$ can now be computed as follows: $$\begin{align}H(\omega)&=\int_{-\infty}^{\infty}\frac{\cos(\omega_0t)}{\pi t}e^{-j\omega t}dt\\&=-j\int_{-\infty}^{\infty}\frac{\cos(\omega_0t)}{\pi t}\sin(\omega t)dt\\&=-\frac{j}{2}\int_{-\infty}^{\infty}\frac{\sin[(\omega+\omega_0)t]+\sin[(\omega-\omega_0)t]}{\pi t}dt\\&=-\frac{j}{2}\big[\textrm{sgn}(\omega+\omega_0)+\textrm{sgn}(\omega-\omega_0)\big]\\&=\begin{cases}-j,&\omega>\omega_0\\0,&-\omega_0<\omega<\omega_0\\j,&\omega<-\omega_0\end{cases}\end{align}$$
This means that $H(\omega)$ is an ideal Hilbert transformer with a high pass characteristic. As shown in the last equation of Dan's answer, since the given impulse response must equal the concatenation of an ideal high pass filter with an ideal Hilbert transformer, i.e., $$\mathscr{H}\left\{\delta(t)-\frac{\sin(\omega_0t)}{\pi t}\right\}=\frac{\cos(\omega_0t)}{\pi t}$$ and because $$\mathscr{H}^2\{x(t)\}=-x(t)$$ we obtain the following side results: $$\mathscr{H}\left\{\frac{\sin(\omega_0t)}{\pi t}\right\}=\frac{1-\cos(\omega_0t)}{\pi t}$$ and $$\mathscr{H}\left\{\frac{\cos(\omega_0t)}{\pi t}\right\}=\frac{\sin(\omega_0t)}{\pi t}-\delta(t)$$
Naturally, there are several ways of computing the Fourier transform of the given impulse response. Summarizing, three ways that are rather straightforward are:
1. Solving the Fourier integral directly using identity $(1)$, as shown above.
2. Realizing that the given impulse response is a modulated version of the impulse response of an ideal Hilbert transformer, as explained in Dan's answer.
3. Starting with the well-known Fourier transform of $\cos(\omega_0t)$ and noting that dividing by $t$ corresponds to integration (times $-j$) in the frequency domain. The integration constant is easily found by requiring that the resulting transform is odd (and purely imaginary), as must be the case for a real-valued odd time domain function.
