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?

We can write $a\sin x+b\cos x=C \sin(x+y)$ where $C=\sqrt{a^2+b^2}$ and $y=\arctan (b/a)$ with the arctangent function accounting for the individual signs of $a$ and $b$ (see Note).  

Thus, the maximum is $C=\sqrt{a^2+b^2}$.


NOTE:
We can see that this is correct using only the addition angle formula for the sine function along with the identity $\sin^2 x+\cos^2 x=1$.  Write
$$C\sin(x+y)=(C\cos y)\sin x+(C\sin y)\cos x \implies a=C\cos y \,\,\text{and,}\,\, b=C\sin y$$  
Solving for $C$ we have $C^2=a^2+b^2\implies C=\sqrt{a^2+b^2}$.  
Solving for $-\pi\le y\le \pi$ we have $\tan y=b/a \implies y=\arctan (b/a)$ for $a>0$, $y=\arctan(b/a)+\pi$ for $a<0$ and $b>0$, and $y=\arctan(b/a)-\pi$ for $a<0$ and $b<0$.
