Trigonometric Inequalities

In one sentence

Trigonometric inequalities come in two kinds. To solve one, you “see” the solution set on the graph and the unit circle. To prove one, you often turn trig values into geometric quantities you can compare (areas, slopes, heights). For inequalities in a triangle there is one more very useful move, step-by-step adjustment: keep the sum of the angles fixed and even them out one step at a time.

Why does evening out the angles make $sin A + sin B + sin C$ larger? Because on $[0, π]$ the sine curve is a concave arch (it bulges upward): the centroid of any points on the arch lies below the arch.

Adjustment: sin A + sin B + sin C ≤ 3√3/2

Step 1 / 4

Put the three angles on the arch
$A = 3 0^{\circ}$ , $B = 5 0^{\circ}$ , $C = 10 0^{\circ}$ . The three points $(x_{i}, sin x_{i})$ are all on the arch, and the triangle they form lies entirely below it.

$sin 3 0^{\circ} + sin 5 0^{\circ} + sin 10 0^{\circ} \approx 2.251$
The centroid is below the arch
Adjust one step
One more step and they are all equal

Stop 1 · Solving inequalities: look within one period

Predict first

In $[0, 2 π)$ , how many intervals make up the solution set of $cos x > \frac{1}{2}$ ?

123

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

The thick parts are where the inequality holds. Switch between

sin

cos

and

tan

, and drag

a

Stop 2 · Comparing sizes: turn them into areas

For $0 < x < \frac{π}{2}$ there is a basic inequality, $sin x < x < tan x$ . Its proof draws all three numbers as areas.

sin x < x < tan x

Step 1 / 4

The small triangle
$P$ is on the unit circle, $O A = 1$ , and the distance from $P$ to $O A$ is $sin x$ .

$S_{△ O A P} = \frac{1}{2} sin x$
The sector
The big triangle
One inside the next

Stop 3 · Extreme values in a triangle: two routes

One problem, several solutions

In $△ A B C$ , prove that $sin A + sin B + sin C \leq \frac{3 3}{2}$ .

Step-by-step adjustment

Fix $C$ first. $sin A + sin B = 2 sin \frac{A + B}{2} cos \frac{A - B}{2} \leq 2 cos \frac{C}{2}$ , with equality when $A = B$ .

So it is enough to prove $2 cos \frac{C}{2} + sin C \leq \frac{3 3}{2}$ , a one-variable problem. Or do what the lab above does: each time, replace a pair of angles by “the mean + what is left”. Within two steps all three angles become $6 0^{\circ}$ , and the sum never decreases.

CostEach step needs a sum-to-product, but it is completely elementary.

Bounding by a tangent line

The tangent to the arch at $x = \frac{π}{3}$ is $y = \frac{3}{2} + \frac{1}{2} (x - \frac{π}{3})$ . The arch is concave, so the whole arch lies below this tangent:

$sin x \leq \frac{3}{2} + \frac{1}{2} (x - \frac{π}{3}), 0 \leq x \leq π .$

Put $x = A, B, C$ and add the three inequalities. The brackets on the right add up to $A + B + C - π = 0$ , which gives $sin A + sin B + sin C \leq \frac{3 3}{2}$ .

CostDone in one line, but you have to guess first that equality is at 60°.

Practice

Exercise 1Basic

In $[0^{\circ}, 36 0^{\circ})$ , the values of $x$ with $cos x < \frac{1}{2}$ form the interval $(α^{\circ}, β^{\circ})$ . Find $β - α$ .

Exercise 2Basic

Putting $x = \frac{π}{6}$ into $sin x < x$ gives a lower bound for $π$ . What is this lower bound?

Exercise 3Advanced

Find the maximum of $cos A + cos B + cos C$ in $△ A B C$ .

Exercise 4Challenge

In an acute triangle $A B C$ , let $m$ be the minimum of $tan A + tan B + tan C$ . Find $m^{2}$ .

In one sentence

Adjustment: sin A + sin B + sin C ≤ 3√3/2

Stop 1 · Solving inequalities: look within one period

Stop 2 · Comparing sizes: turn them into areas

sin x < x < tan x

Stop 3 · Extreme values in a triangle: two routes

Step-by-step adjustment

Bounding by a tangent line

Practice

Mistake Museum · Exhibits from this stop

Solving sin x > ½ as if it were a linear inequality