Maximising & Minimising Trig Expressions

To maximise or minimise $a\cos x + b\sin x$, first convert it to auxiliary angle form. The process is:

1. Write $a\cos x + b\sin x = R\cos(x - \alpha)$ and expand the right side.
2. Match coefficients: $R\cos\alpha = a$ and $R\sin\alpha = b$.
3. Square and add: $R^2 = a^2 + b^2$, so $R = \sqrt{a^2+b^2}$ (take positive root).
4. Divide: $\tan\alpha = b/a$, then determine the quadrant of $\alpha$ from the signs of $a$ and $b$.

Example — find the max and min of $3\cos x + 4\sin x$:

$R = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$

$\tan\alpha = \dfrac{4}{3}$, with $a = 3 > 0$ and $b = 4 > 0$, so $\alpha$ is in the first quadrant: $\alpha = \arctan\!\left(\tfrac{4}{3}\right) \approx 53.13°$.

Therefore $3\cos x + 4\sin x = 5\cos(x - \alpha)$.

a x + b x = R(x-) where R = a^2+b^2 and = b/a; Max value = +R; occurs when x = (or + 360°k)

Pause — copy the maximum-value result into your book: $a\sin x + b\cos x = R\sin(x-\alpha)$ has maximum $+R$ at $x = \alpha$ and minimum $-R$ at $x = \alpha+180°$.

We just saw that $a\sin x + b\cos x = R\sin(x-\alpha)$ achieves its maximum $+R$ at $x = \alpha$. That raises a question: what if the domain is restricted so that $x = \alpha$ is not included — how do you find the maximum on the restricted interval? This card answers it → check whether $\alpha$ is in the domain; if not, evaluate $f$ at both endpoints and compare.

When a domain restriction is given (e.g. $0 \leq x \leq \pi$), you must check whether the unrestricted maximum $x = \alpha$ lies inside the allowed domain.

• If $\alpha$ is inside the domain, the maximum value is still $R$.
• If $\alpha$ is outside the domain, evaluate the expression at the two boundary values $x = 0$ and $x = \pi$ (or whatever the endpoints are), and pick the larger.

Example: Find the maximum of $3\cos x + 4\sin x$ for $\pi \leq x \leq 2\pi$.

We found $\alpha \approx 53.13°$, which is NOT in $[\pi, 2\pi]$. So we check endpoints:

• $x = \pi$: $3\cos\pi + 4\sin\pi = -3 + 0 = -3$
• $x = 2\pi$: $3\cos 2\pi + 4\sin 2\pi = 3 + 0 = 3$

Also check $x = \alpha + 180° \approx 233.13°$ (minimum location, inside the domain): value $= -5$.

Maximum on $[\pi, 2\pi]$ is $\mathbf{3}$, achieved at $x = 2\pi$.

Restricted domain checklist: (1) find , (2) is in the domain? (3) if not, check endpoints; Also check + 180° — that is the minimum location

Pause — copy the restricted-domain checklist into your book: (1) find $\alpha$; (2) is $\alpha$ in the domain?; (3) if not, compare $f$ at the endpoints of the domain to determine the actual maximum.