☰ Contents · Mathematics

Trigonometric inequalities and graph transformations

Lessons 31–32 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 10, Parts 1 and 2, 1st edition. EXTREMUM PRESS, Tashkent, 2017
31

The simplest trigonometric inequalities

Textbook, Part 2: pp. 44–47
GoalSolve the simplest inequalities comparing sin x, cos x and tan x with a number a, using the unit circle.
New words
trigonometric inequality · trigonometrik tengsizlikarc of the unit circle · birlik aylana yoyiperiod · davrinterval · oraliq
Explanation

Inequalities of the form a₁ < sin x < b₁, a₂ < cos x < b₂, a₃ < tan x < b₃ are the simplest trigonometric inequalities. The value sin x is the ordinate of a point on the unit circle and cos x is its abscissa. To solve sin x > a we find the arc of points with ordinate greater than a, taking its ends from the equation sin x = a. For cos x > a we take the arc whose abscissa exceeds a. We solve first within one period (say [0, 2π]) and then add the period: 2πk for sin and cos, πk for tan, k ∈ ℤ. If the inequality is strict the endpoints are excluded; if not, they are included. To check, pick one point of the solution and substitute it into the inequality.

Worked examples
sin x ≥ √3/2 (within one period): the ordinate is at least √3/2 for x between π/3 and 2π/3. In general x ∈ [π/3 + 2πk, 2π/3 + 2πk], k ∈ ℤ.
cos x < −1/2: cos x = −1/2 at x = 2π/3 and x = 4π/3. The arc with abscissa below −1/2 lies between these points. Answer: x ∈ (2π/3 + 2πk, 4π/3 + 2πk), k ∈ ℤ.
Class activity

Draw a circle in your notebook, shade the arc satisfying sin x > 1/2 with a coloured pencil and write the angles of its ends; your partner does the same for cos x ≤ 1/2.

Practice
1
What is the solution of sin x > 0 on [0, 2π]?
2
Solve cos x > 0 on [0, 2π].
3
Solve sin x < −1/2 on [0, 2π].
4
Why does sin x ≤ 2 hold for every x?