Finding amplitude of oscillation So I know how to find the amplitude of oscillation.  It's just the coefficient of the trig function. In general it would be $$ x = A\sin\left(\omega t +\phi_0\right) $$ and $A$ would be the amplitude. But my problem is asking for the amplitude of $$ x=-\cos(t)+3\sin\left(t-\frac{\pi}{6}\right) $$I don't exactly know what to do if there are 2 trig functions. The amplitude is $3.606$ but I don't know how to get that.
 A: Expanding the explanations  given in the comments by Artem and André Nicolas to the question. 
We can always find constants $A,B$ and $C$ such that the equation
\begin{equation*}
x(t)=A\cos \omega t+B\sin \omega t
\end{equation*}
is identical to the equation
\begin{equation*}
x(t)=C\sin \left( \omega t+\phi _{0}\right) =\left( C\sin \phi _{0}\right)
\cos \omega t+\left( C\cos \phi _{0}\right) \sin \omega t.
\end{equation*}
To expand $C\sin \left( \omega t+\phi _{0}\right) $ we applied the
trigonometric identity 
\begin{equation*}
\sin (a+b)=\sin a\cos b+\cos a\sin b.
\end{equation*}
Equating the coefficients of $\cos \omega t$ and $\sin \omega t$ gives
\begin{equation*}
A=C\sin \phi _{0},\qquad B=C\cos \phi _{0},
\end{equation*}
while squaring and adding these last equations gives
\begin{equation*}
C=\sqrt{A^{2}+B^{2}}\geq 0.
\end{equation*}
Dividing one by the other yields 
\begin{equation*}
\tan \phi =\frac{A}{B}.
\end{equation*}
In this case we have that
\begin{equation*}
x(t)=C\sin \left( \omega t+\phi _{0}\right) =-\cos t+3\sin (t-\frac{\pi }{6}
)=-\frac{5}{2}\cos t+\frac{3\sqrt{3}}{2}\sin t.
\end{equation*}
We changed the notation of the amplitude of the wave $x(t)$ to $C$
instead of $A\ $as in the question. We see that $\omega =1$. The expansion
of $\sin (t-\frac{\pi }{6})$ follows from the identity
\begin{equation*}
\sin (a-b)=\sin a\cos b-\cos a\sin b
\end{equation*}
and the trigonometric values
\begin{equation*}
\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\quad \sin \frac{\pi }{6}=\frac{1}{2}.
\end{equation*}
From the numeric values
\begin{equation*}
A=-\frac{5}{2},\qquad B=\frac{3\sqrt{3}}{2},
\end{equation*}
we find that the amplitude is 
\begin{equation*}
C=\sqrt{A^{2}+B^{2}}=\sqrt{13}\approx  3. 6056,
\end{equation*}
the same value of yours. If we wanted to compute the phase angle $\phi _{0}\ $we would get
\begin{eqnarray*}
\tan \phi _{0} &=&\frac{A}{B}=\frac{-5\sqrt{3}}{9}, \\
\phi _{0} &=&\arctan \frac{-5\sqrt{3}}{9}\approx -0.766\,16\text{ }\mathrm{
rad}\approx -43.898{{}^\circ}.
\end{eqnarray*}
Graph of 
\begin{equation*}
x(t)=\sqrt{13}\sin \left( t-0.766\,16\right) 
\end{equation*}

