We can directly use the fact that a Triangular function (same as one in your question) is a result when two rectangular pulses(with same width) gets convoluted...
Now defining a rectangular function
$$A.\Pi \left(\frac{t}{\tau}\right)=\left\{\begin{matrix}A~,-\tau/2<t<\tau/2
\\ 0~,~ ~~~~~~~\text{otherwise}\end{matrix}\right.$$
Now it's FT is fiven by $$\mathcal{F}\left\{A.\Pi \left(\frac{t}{\tau}\right)\right\}=A\tau\frac{\sin(\omega\tau/2)}{\omega\tau/2}=\mathrm{sa}\left(\frac{\omega\tau}{2}\right)=A\tau. \mathrm{sinc}(f\tau)$$
And when two of these pulses are convoluted we can obtain a triangular pulse
And it's Ft is given by $$\begin{align}\mathcal{F}\left\{A\mathrm{Trig}\left(\frac{t}{\tau}\right) \right\}&=A\tau.\frac{\sin^2(\omega \tau/2)}{\omega^2\tau^2/4}\\&=A\tau.\mathrm{Sa}^2 \left(\frac{\omega \tau }{2}\right)\\&=A\tau .\mathrm{Sinc}^2(f \tau)\end{align}$$
In this case, we've got $A=1$ and $\tau=1$ , so required FT is
$$\begin{align}\mathcal{F}\left\{\mathrm{Trig}\left(t\right) \right\}&=\mathrm{Sa}^2 \left(\frac{\omega}{2}\right)\\&=\frac{\sin^2 (\omega/2)}{\omega^2/4}\\&=\mathrm{Sinc}^2(f)\\&=\mathrm{Sinc}^2 \left(\frac{\omega}{2\pi}\right)\end{align}$$
Hope this helps, please tell me if by any means i can improve my answer , thanks !