Understanding Maximum and Minimum Values of Trigonometric Functions

The maximum and minimum values of trigonometric functions refer to the greatest and least values attained by standard trigonometric functions such as sine, cosine, tangent, and their reciprocals, over their domains. The determination of these extreme values is governed by the functional structure and ranges of these periodic functions.

Algebraic Structure and Ranges of Basic Trigonometric Functions

Sine function: $y = \sin x$ is defined for all real $x,$ and its range is $[-1, 1].$

This means for every $x \in \mathbb{R},$ $-1 \leq \sin x \leq 1.$ Therefore, $\sin x$ attains its maximum value $1$ and minimum value $-1.$ Maximum occurs for $x = 2n\pi + \frac{\pi}{2},$ minimum for $x = 2n\pi + \frac{3\pi}{2},$ with $n \in \mathbb{Z}.$

Cosine function: $y = \cos x$ is defined for all real $x,$ with range $[-1, 1].$

Every $x \in \mathbb{R}$ satisfies $-1 \leq \cos x \leq 1,$ so $\cos x$ reaches maximum $1$ for $x = 2n\pi,$ and minimum $-1$ for $x = 2n\pi + \pi,$ with $n \in \mathbb{Z}.$

Tangent function: $y = \tan x$ is defined wherever $x \neq (2n + 1)\frac{\pi}{2},$ and its range is $(-\infty, \infty).$

The tangent function is unbounded and attains arbitrarily large positive and negative values as $x$ approaches its points of discontinuity (odd multiples of $\frac{\pi}{2}$). Thus, $\tan x$ does not have finite maximum or minimum values.

Cotangent function: $y = \cot x$ is defined wherever $x \neq n\pi,$ and its range is $(-\infty, \infty).$

Similar to tangent, cotangent is unbounded near its points of discontinuity (integer multiples of $\pi$) and has no finite maximum or minimum.

Secant function: $y = \sec x$ is defined wherever $x \neq (2n+1)\frac{\pi}{2}$ with range $(-\infty, -1]\cup [1, \infty)$.

The minimum value in $[1, \infty)$ is $1$ (at $x = 2n\pi$) and the maximum is unbounded as $x \to (2n+1)\frac{\pi}{2}^{-}$. In $(-\infty, -1],$ the minimum value is unbounded and the maximum value is $-1$ (at $x = (2n+1)\pi$).

Cosecant function: $y = \csc x$ is defined wherever $x \neq n\pi$ with range $(-\infty, -1] \cup [1, \infty)$.

For $y = \csc x,$ the minimum in $[1, \infty)$ is $1$ (at $x = n\pi + \frac{\pi}{2}$), and the maximum in $(-\infty, -1]$ is $-1$ (at $x = n\pi - \frac{\pi}{2}$). Both positive and negative branches are unbounded approaching points of discontinuity.

Maximum and Minimum of Linear Combinations: $a \sin x + b \cos x$

Result: For real numbers $a$ and $b,$ $a \sin x + b \cos x$ attains maximum $\sqrt{a^2 + b^2}$ and minimum $-\sqrt{a^2 + b^2}$.

To establish this, consider $y = a \sin x + b \cos x.$ This can be rewritten as $y = R \sin(x + \alpha),$ where $R = \sqrt{a^2 + b^2}$ and $\alpha = \arctan\left(\frac{b}{a}\right),$ given $a \neq 0.$

First, compute $R = \sqrt{a^2 + b^2}$.

Next, write:

$a \sin x + b \cos x = \sqrt{a^2+b^2} \left( \frac{a}{\sqrt{a^2+b^2}} \sin x + \frac{b}{\sqrt{a^2+b^2}} \cos x \right )$

Let $\cos \alpha = \dfrac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \dfrac{b}{\sqrt{a^2+b^2}}$.

Substitute these into the expression:

$a \sin x + b \cos x = \sqrt{a^2+b^2} (\sin x \cos \alpha + \cos x \sin \alpha)$

From the angle addition formula: $\sin x \cos \alpha + \cos x \sin \alpha = \sin(x+\alpha)$.

Thus, $a \sin x + b \cos x = \sqrt{a^2+b^2}\, \sin(x+\alpha)$.

$\sin(x+\alpha)$ attains maximum $1$ and minimum $-1.$

Thus, maximum of $a \sin x + b \cos x$ is $\sqrt{a^2+b^2} \cdot 1 = \sqrt{a^2+b^2}$ and minimum is $\sqrt{a^2+b^2} \cdot (-1) = -\sqrt{a^2+b^2}$.

For a closely related concept involving quadratic expressions, refer to Maximum And Minimum Value Of A Quadratic Polynomial.

Explicit Determination of Extrema: $a \sin x + b \cos x + c$

The expression $a \sin x + b \cos x + c$ achieves maximum $c + \sqrt{a^2 + b^2}$ and minimum $c - \sqrt{a^2 + b^2},$ since it is vertically shifted by $c$.

Maximum and Minimum for Quadratic Trigonometric Forms: $a \sin^2 x + b \sin x + c$

Let $y = a \sin^2 x + b \sin x + c.$ Substitute $t = \sin x$ where $-1 \leq t \leq 1.$ Then, $y = a t^2 + b t + c.$

This quadratic in $t$ attains its maximum or minimum at the endpoints $t = -1, 1,$ or at the vertex $t_0 = -\frac{b}{2a}$ if $t_0 \in [-1, 1].$

Evaluate $y_{-1} = a(-1)^2 + b(-1) + c = a - b + c,$

Evaluate $y_{1} = a(1)^2 + b(1) + c = a + b + c.$

Compute vertex $t_0 = -\dfrac{b}{2a}.$ If $-1 \leq t_0 \leq 1,$ compute $y_0 = a t_0^2 + b t_0 + c$ as well.

The maximum is the largest among $y_{-1}, y_1, y_0$ (if $t_0$ is in $[-1,1]$), and the minimum is the smallest among them.

Exact Determination by Calculus: Critical Points

For trigonometric functions of the form $f(x),$ to determine absolute maxima or minima in a given interval, compute derivatives, set $f'(x) = 0,$ solve for critical points within the interval, and substitute into $f(x)$ along with endpoints.

For instance, the maximum of $\sin x$ in $[0, 2\pi]$ occurs at $x = \frac{\pi}{2}$, $f(\frac{\pi}{2}) = 1,$ and minimum at $x = \frac{3\pi}{2},$ $f(\frac{3\pi}{2}) = -1.$

Extrema of Inverse Trigonometric Functions

The inverse functions $\arcsin x, \arccos x, \arctan x$ have the following ranges: $\arcsin x \in [-\frac{\pi}{2}, \frac{\pi}{2}], \; \arccos x \in [0, \pi], \; \arctan x \in (-\frac{\pi}{2}, \frac{\pi}{2}).$ The minimum and maximum for each occur at the endpoints of their domain $x \in [-1, 1]$ for $\arcsin$ and $\arccos$ and $x \in \mathbb{R}$ for $\arctan$.

Worked Example: Maximum Value of $3\sin x - 4\cos x$

Example: Find the maximum and minimum values of $y = 3\sin x - 4\cos x$.

Given $a = 3, \; b = -4.$ Compute $R = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5.$

Rewrite $y = 5\sin(x + \alpha)$ with $\tan\alpha = \dfrac{-4}{3}.$

Maximum value of $y = 5$, minimum value is $-5$.

Common Errors on Maximum and Minimum of Trigonometric Expressions

Common Error: Failing to account for domain restrictions of expressions involving secant, cosecant, tangent, or cotangent can lead to non-existent extrema or incorrect answers.