The problem is to find the maximum height of the curve $y=3\cos{x-4 \sin{x}}$ above the $x$-axis. The way I'd go about it is by finding the derivative of $y$ and then setting it equal to zero. However, I happened to find the following solution on the internet, which yields the same answer but without using Calculus:
$$3\cos{x}-4\sin{x}=5\left(\frac{3}{5}\cos{x}-\frac{4}{5}\sin{x}\right)=5·\sin({\phi}-x)$$ $$\max(5·\sin{(\phi-x)})=5$$ $$\sin{\phi}=\frac{3}{5}$$
I am a bit confused though. Apart from the first inequality, what exactly is being done in each line? I know the difference formula for sine, but how does one go from $3/5\cos{x}-4/5\sin{x}$ to $\sin({\phi-x})$? I'd really appreciate it if someone could shed some light on this. Thanks.
Let $\phi$ be the acute angle such that $\sin\phi=3/5$. Then, $\cos\phi=\sqrt{1-(3/5)^2}=4/5$ and $$\sin(\phi-x)=\sin\phi\cos x-\cos \phi\sin x=\frac35\cos x-\frac45\sin x$$
Then the maximum of $5\sin(\phi-x)$ comes when $\sin(\phi-x)=1$ and it is $5$.
In general, if $a,b\in\Bbb R$, we can write $$a\sin x+b\cos x=\sqrt{a^2+b^2}\sin(\alpha+x)$$ for some angle $\alpha$. This angle holds $\sin\alpha=a/\sqrt{a^2+b^2}$ and $\cos\alpha=b/\sqrt{a^2+b^2}$.