Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$

Solution 1:

You are largely correct. However you could prove the first part in a very simple way:

We can write $$P= a\cos x+b\sin x =\sqrt {a^2+b^2} \left[\frac {a}{\sqrt {a^2+b^2}}\cos x+\frac{b}{\sqrt {a^2+b^2}}\sin x \right] $$ Now we can take $\frac{a}{\sqrt {a^2+b^2}} $ as $\cos \alpha $ giving us $$P=\sqrt {a^2+b^2}[\cos \alpha \cos x+\sin \alpha \sin x]=\sqrt{a^2+b^2}\cos (x-\alpha) $$ And also $$\alpha =\arccos \frac {a}{\sqrt {a^2+b^2}} $$ which I think maybe a small typo in your calculation.

You need not prove the first part with the help of a triangle but your approach is fine. Hope it helps.

Solution 2:

The proof is only valid for $0<x<\frac\pi2$. For a more general proof, let: $$a\cos x+b\sin x\ \equiv\ R\cos(x-\alpha)=R\cos\alpha\cos x + R\sin\alpha\sin x$$ So $a=R\cos\alpha$ and $b=R\sin\alpha$, from which you can readily find $R$ and $\alpha$.