Two Important Inequalities

Inverse Trigonometric FunctionsLimits And Continuity

In this section, we establish two fundamental trigonometric inequalities for all angles θ measured in radians. These inequalities bound sin ⁡ θ and 1 − cos ⁡ θ near the origin and form the geometric foundation for critical limit proofs in calculus.

Quick Reference

Inequality Domain Primary Application
− < / t d >< t d > θ < / t d >< t d >≤ sin ⁡ θ ≤< / t d >< / t r >< t r >< t d > - \theta \le 1 - \cos\theta \le

Statement of the Inequalities

In this section, we will establish the following inequalities for all angles θ measured in radians (see the following figure):

− | θ | ≤ sin ⁡ θ ≤ | θ | and − | θ | ≤ 1 − cos ⁡ θ ≤ | θ |
Graphs showing -|theta| <= sin theta <= |theta| and -|theta| <= 1 - cos theta <= |theta| for all theta
For all θ , − | θ | ≤ sin ⁡ θ ≤ | θ | and − | θ | ≤ 1 − cos ⁡ θ ≤ | θ | .

Proof of the Inequalities

Proof

To prove, let's picture θ as an acute angle in standard position (see the following figure). Because the circle is a unit circle, the length of the arc A P ⌢ is r θ = θ (with r = 1 ) and the length of the line segment A P ≤ θ .

Unit circle diagram showing triangles OHP and AHP with arc AP
Unit circle setup for proving trigonometric inequalities.

In the right triangle △ O H P :

H P = O P sin ⁡ θ = 1 ⋅ sin ⁡ θ = sin ⁡ θ O H = O P cos ⁡ θ = 1 ⋅ cos ⁡ θ = cos ⁡ θ

Because O A = 1 = O H + H A = cos ⁡ θ + H A , we have H A = 1 − cos ⁡ θ .

In the right triangle △ A H P :

H P = sin ⁡ θ , H A = 1 − cos ⁡ θ

Using the Pythagorean theorem, we have:

H P 2 + H A 2 = ( length of line segment  A P ) 2 sin 2 ⁡ θ + ( 1 − cos ⁡ θ ) 2 = ( length of line segment  A P ) 2

but we learned that the length of the line segment A P ≤ length of  A P ⌢ = θ . Therefore,

sin 2 ⁡ θ + ( 1 − cos ⁡ θ ) 2 ≤ θ 2

On the left hand side of the above inequality, we have two nonnegative quantities, so each of them is smaller than or equal to their sum and hence each of them is less than or equal to θ 2 . That is,

sin 2 ⁡ θ ≤ θ 2 and ( 1 − cos ⁡ θ ) 2 ≤ θ 2

They are equivalent to:

− | θ | ≤ sin ⁡ θ ≤ | θ | and − | θ | ≤ 1 − cos ⁡ θ ≤ | θ |

[Remember that if | x | ≤ r , then − r ≤ x ≤ r .]

We proved that these inequalities hold when 0 < | θ | < π 2 .

When θ = 0 , these inequalities are obvious since sin ⁡ 0 = 0 and 1 − cos ⁡ 0 = 0 .

For | θ | > π 2 ≈ 1.57 , these inequalities are once again obvious since − 1 ≤ sin ⁡ θ ≤ 1 ≤ | θ | and 0 ≤ 1 − cos ⁡ θ ≤ 2 ≤ | θ | .

Therefore, for all θ :

− | θ | ≤ sin ⁡ θ ≤ | θ | and − | θ | ≤ 1 − cos ⁡ θ ≤ | θ |
Inverse Trigonometric FunctionsLimits And Continuity
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