# How to prove this trigonometric inequality?

I am trying to prove the inequality \begin{aligned} \cos((b - a) t) - \cos((b + a) t) + t \sqrt{n (n - 1)} (a - b)\sin((a + b) t)\\ - t \sqrt{n (n - 1)} (a + b) \sin((b - a) t)\le0 \end{aligned} where $a=\sqrt{k/n}$ and $b=\sqrt{k/(n-1)}$ for $t\in[0,1]$, $k>0$ and $n\ge2$.

After plotting this function, I am convinced that the inequality holds. I tried to prove it using derivative with respect to $t$, but ended up with sines and cosines with coefficients getting more and more complicated. Is there a way to prove this?

$$\cos\left((b-a)t\right)=\cos(bt-at)=\cos(bt)\cos(at)+\sin(bt)\sin(at)$$ $$\cos((b+a)t)=\cos(bt)\cos(at)-\sin(bt)\sin(at)$$ $$\cos\left((b-a)t\right)-\cos((b+a)t)=2\sin(bt)\sin(at)$$ try from here