Trigonometric Substitution and Applications

In one sentence

Some algebraic expressions naturally live on a circle: $x^{2} + y^{2} = r^{2}$ says that the point $(x, y)$ is on a circle, $1 - x^{2}$ is the height of the upper semicircle, and $\frac{2 x}{1 + x ^{2}}$ is the sine of a double angle. Replace $x$ and $y$ by trigonometric functions and the constraint holds automatically. What is left is a trigonometric problem in one variable, which an auxiliary angle can usually finish off.

Example: x² + y² = 4. Find the max and min of x + √3 y

Step 1 / 4

The constraint becomes a parameter
The point $(x, y)$ is on the circle of radius $2$ , so we can write $x = 2 cos t$ , $y = 2 sin t$ . Whatever $t$ is, the constraint holds.

$x = 2 cos t, y = 2 sin t$
The target becomes a trig expression
Maximum: tangency
Minimum: tangency on the other side

Stop 1 · Radicals: the upper semicircle

$1 - x^{2}$ is the height of the upper half of the unit circle. Let $x = cos t$ with $t \in [0, π]$ ; then $1 - x^{2} = sin t$ . Watch the range of $t$ : it is what keeps the point on the upper semicircle.

Example: find the range of y = x + √(1 − x²)

Step 1 / 3

Substitute
$x \in [- 1, 1]$ . Let $x = cos t$ with $t \in [0, π]$ : then $x$ takes every value in $[- 1, 1]$ exactly, and $sin t \geq 0$ .

$y = cos t + sin t = 2 sin (t + \frac{π}{4})$
Maximum
The minimum is at an endpoint

Stop 2 · Fractions: the universal substitution

Let $x = tan t$ with $t \in (- \frac{π}{2}, \frac{π}{2})$ . Then

\frac{2 x}{1 + x ^{2}} = sin 2 t, \frac{1 - x ^{2}}{1 + x ^{2}} = cos 2 t .

$x$ can run off to infinity, yet these two fractions stay locked inside the circle.

Predict first

As $x$ ranges over all real numbers, what is the maximum of $\frac{2 x}{1 + x ^{2}}$ ?

Choose an answer first; then you can see the result.

Stop 3 · Substituting for two expressions at once

For an expression like $\frac{x + y}{1 - x y}$ , putting $x = tan α$ , $y = tan β$ turns it into $tan (α + β)$ . When two constraints $a^{2} + b^{2} = 1$ and $c^{2} + d^{2} = 1$ appear together, give each pair its own angle. Here is the second kind, solved in two ways.

One problem, several solutions

Given $a^{2} + b^{2} = 1$ and $c^{2} + d^{2} = 1$ , find the maximum of $a c + b d$ .

Trigonometric substitution

Let $a = cos α$ , $b = sin α$ , $c = cos β$ , $d = sin β$ . Then

$a c + b d = cos α cos β + sin α sin β = cos (α - β) \leq 1,$

with equality when $α = β$ . Geometrically, $a c + b d$ is the cosine of the angle between two unit vectors.

CostYou need two angles, but the conclusion is plain to see.

The mean inequality

$a c \leq \frac{a ^{2} + c ^{2}}{2}$ and $b d \leq \frac{b ^{2} + d ^{2}}{2}$ ; adding gives $a c + b d \leq \frac{1 + 1}{2} = 1$ , with equality when $a = c$ and $b = d$ .

CostOne line, but you have to spot the pairing.

Practice

Exercise 1Basic

Real numbers $x$ , $y$ satisfy $x^{2} + y^{2} = 1$ . Find the maximum of $3 x + 4 y$ .

Exercise 2Advanced

Find the minimum of $y = x + 4 - x^{2}$ .

Exercise 3Advanced

$x$ ranges over all real numbers. Find the maximum of $\frac{1 - x ^{2} + 2 3 x}{1 + x ^{2}}$ .

Exercise 4Challenge

Real numbers $x$ , $y$ satisfy $x^{2} + y^{2} = 1$ . Find the minimum of $(x + 1) (y + 1)$ .

In one sentence

Example: x² + y² = 4. Find the max and min of x + √3 y

Stop 1 · Radicals: the upper semicircle

Example: find the range of y = x + √(1 − x²)

Stop 2 · Fractions: the universal substitution

Stop 3 · Substituting for two expressions at once

Trigonometric substitution

The mean inequality

Practice

Mistake Museum · Exhibits from this stop

Forgetting to restrict the angle after substituting