<s>[INST] 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
 [/INST] You can proceed with your initial path as follows
$$a\cos x +b\sin  x$$
Multiply and divide by $\displaystyle\frac{1}{\sqrt{a^2+b^2}}$
$$\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\cos x +\frac{b}{\sqrt{a^2+b^2}}\sin  x)$$
Now $(\frac{a}{\sqrt{a^2+b^2}})^2+(\frac{b}{\sqrt{a^2+b^2}})^2=1$ and both terms are less than $1$
Therefore if we let one as $\cos \phi $ other can be let us $\sin \phi$
Therefore we have
$$\sqrt{a^2+b^2}(\cos\phi\cos x + \sin\phi\sin x)$$
$$A\cos (\phi-x)$$
Thus there exists two reals $A=\sqrt{a^2+b^2}$ and $\phi= \cos^{-1}(\frac{a}{\sqrt{a^2+b^2}})$ for which $a\cos x +b\sin  x$ is identically equal to $A\cos (\phi-x)$
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