Consecutive company

We can write

$3 \sin \theta + 4\cos \theta \equiv r\sin (\theta + \alpha)=r\sin\theta\cos\alpha+r\cos\theta\sin\alpha$

where $r \ge 0$ and $0 \leq \alpha \leq \frac{\pi}{2}$

Thus,

$r\cos\alpha=3$

and

$r\sin\alpha=4$

Therefore,

$3^2+4^2=r^2(\cos^2\alpha+\sin^2\alpha) \rightarrow r=5$

Hence, $3\sin\theta+4\cos\theta=5\sin(\theta+\alpha)$ .

and from $-1 \leq \sin(\theta+\alpha) \leq 1$ ,

we have

$-3 \leq 5\sin(\theta+\alpha)+2 \leq 7$

Hence, $a+b=7-3=4$

(By the way, $\alpha=\tan^{-1}(\frac{r\sin\alpha}{r\cos\alpha})=\tan^{-1}(\frac{4}{3})=0.9273rad$ )

We can think of this problem as an AC circuit analysis problem. Imagine a series circuit with three voltage sources in series. The phasor representation of $3\sin\theta$ is $3 \angle{0}$ and $4\cos\theta$ is $4 \angle{90}$ . The sum of these two phasors is $5 \angle {53.13}$ . This implies that the sum of the two sinusoids is another sinuosid with minimum of $-5$ and maximum of $5$ . Adding up the dc source (superposition principle) gives you the range $[-3,7]$ .

Let's try a calculus approach here. Let $f(\theta) = 3\sin(\theta) + 4\cos(\theta) + 2$ such that its first & second derivatives compute to:

$f'(\theta) = 3\cos(\theta) - 4\sin(\theta),$

$f''(\theta) = -3\sin(\theta) - 4\cos(\theta).$

Taking $f'(\theta) = 0$ yields $\tan(\theta) = \frac{3}{4}$ . Since the tangent is positive in the first and third quadrants of the $xy-$ plane, this results in the angles $\theta = 36.87^{\circ}, 216.87^{\circ}$ . If we evaulate $f''(\theta)$ at these two critical angles, we obtain:

$f''(36.87^{\circ}) < 0$ since the sine and cosine are both positive in the first quadrant $\Rightarrow$ MAXIMUM.

$f''(216.87^{\circ}) > 0$ since the sine and cosine are both negative in the third quadrant $\Rightarrow$ MINIMUM.

Returning to our original function, we now calculate its range:

$f(36.87^{\circ}) = 7$ and $f(216.87^{\circ}) = -3$

or $[a,b] = [-3,7] \Rightarrow a+b = \boxed{4}.$