trigonometric inequality bound? Suppose that one wants to determine an upper bound of the trigonometric expression
$$
a\sin(x)+b\cos(x),
$$
where $a,b\in\mathbb{R}$. My instinct is to proceed as follows:
$$
a\sin(x)+b\cos(x)\leq |a|+|b|,
$$
which is correct. If one were to drop the modulus signs, would that make the bound incorrect?

the conservative(naive) upper bound $|a| + |b|$ is too big because $\sin x$  and $\cos x$ cannot be both max or both min simultaneously.  the reason is the ever present constraint $$\sin^2 x +\cos^2 x = 1.$$
what happens is $$\begin{align}a\sin x  + b \cos x &= \sqrt{a^2 + b^2}\left(\frac a{\sqrt{a^2 + b^2}}\sin x + \frac b{\sqrt{a^2 + b^2}}\cos x\right) \\ 
&=\sqrt{a^2+b^2}\cos(x-t), \cos t = \frac b{\sqrt{a^2 + b^2}}, \sin t = \frac a{\sqrt{a^2 + b^2}}\end{align}$$
therefore $$\lvert a\cos x + b\sin x \rvert \le \sqrt{a^2 + b^2.}$$
