Simplifiying $A \cos(x-\phi)$ I have to proof that we can transform this expression : $$a \cos x + b \sin x$$
to this one :  $$A \cos (x-\phi)$$
Indeed, I can do the reverse path using the formula of $\cos(x-y)= \cos x \cos y + \sin x \sin y$
but I get stuck on the initial path.
Which trig formula should I use?
Thanks

You have
\begin{align*}
A\cos(x-\phi)&=A\cos x\cos \phi+A\sin x\sin \phi\\
&=(A\cos \phi)\cos x+(A\sin \phi)\sin x\\
&=a\cdot \cos x+ b\cdot \sin x
\end{align*}
where $a=A\cos \phi$ and $b=A\sin \phi$
Now, let us try to solve the equations
$$a=A\cos \phi\\ b=A\sin \phi$$
Squaring the equations and adding them, we get
$$a^2+b^2=A^2\cos^2\phi+A^2\sin^2\phi=A^2\left(\cos^2\phi+\sin^2\phi\right)=A^2$$
which gives
$$A=\sqrt{a^2+b^2}$$
Put this back into the first equation and see that
$$a=\sqrt{a^2+b^2}\cos \phi$$
which means
$$\phi=\cos^{-1}\frac a {\sqrt{a^2+b^2}}$$
So, we have proved that given the expression
$$a\cos x+b\sin x$$
we can write it as
$$\sqrt{a^2+b^2}\cos\left(x-\cos^{-1}\frac a {\sqrt{a^2+b^2}}\right)$$
which gives you the required form.
