I'm preparing notes for teaching next week's class in differential equations. My hand calculation solution of $$y''+192y=0, y(0)=1/6, y'(0)=-1$$ is $$y=\frac16\cos 8\sqrt3 t-\frac{\sqrt3}{24}\sin 8\sqrt3 t$$ Now a few more hand calculations provides this equivalent solution: $$y=\frac{\sqrt{19}}{24}\cos(8\sqrt3 t-\tan^{-1}(-\sqrt3/4))$$
Now, when we solve the equation using DSolveValue, we get the following result.
sol = DSolveValue[{u''[t] + 192 u[t] == 0, u[0] == 1/6, u'[0] == -1}, u[t],
t]
Out[276]= 1/24 (4 Cos[8 Sqrt[3] t] - Sqrt[3] Sin[8 Sqrt[3] t])
Which agrees with my hand calculated solution. Moreover, I was able to check that my final form was equivalent to the output provided by DSolveValue.
sol - Sqrt[19]/24*Cos[8 Sqrt[3] t - ArcTan[-Sqrt[3]/4]] // Simplify
Out[293]= 0
Now my question. Is there a simple way in Mathematica to convert the expression
$$\frac16\cos 8\sqrt3 t-\frac{\sqrt3}{24}\sin 8\sqrt3 t$$
to
$$\frac{\sqrt{19}}{24}\cos(8\sqrt3 t-\tan^{-1}(-\sqrt3/4))$$
These are the hand calculations used to make the change.