We prove some results that are needed in the application of calculus to trigonometry.

**Theorem:** If θ is a positive angle, less than a right angle (expressed in radians), then 0 < sin(θ) < θ < tan(θ).

**Proof:** Consider a circle, centre O, radius r, and choose two points A and B on the circumference such that angle AOB is less than a right angle. Draw a tangent to the circle at B, and let OA produced intersect it at C. Clearly

- 0 < area(Δ OAB) < area(sector OAB) < area(Δ OBC)

i.e.

- 0 <
^{1}⁄_{2}r^{2}sin(θ) <^{1}⁄_{2}r^{2}θ <^{1}⁄_{2}r^{2}tan(θ)

and the result follows.

**Corollary:** If θ is a negative angle, more than minus a right angle (expressed in radians), then 0 > sin(θ) > θ > tan(θ). (This follows from sin(-θ) = -sin(θ) and tan(-θ) = -tan(θ).)

**Corollary:** If θ is a non-zero angle, less than a right angle but more than minus a right angle (expressed in radians), then 0 < |sin(θ)| < |θ| < |tan(θ)|.

**Theorem:** As and .

**Proof:** Dividing the result of the previous theorem by sin(θ) and taking reciprocals,

- .

But cos(θ) tends to 1 as θ tends to 0, so the first part follows.

Dividing the result of the previous theorem by tan(θ) and taking reciprocals,

- .

Again, cos(θ) tends to 1 as θ tends to 0, so the second part follows.

**Theorem:** If θ is as before then .

**Proof:**

- .

**Theorem:** If θ is as before then .

**Proof:**

- .

- .

- .

**Theorem:** sin(θ) and cos(θ) are continuous functions.

**Proof:** For any *h*,

- ,

since |cos(x)| cannot exceed 1 and |sin(x)| cannot exceed |x|. Thus, as

- ,

proving continuity. The proof for cos(θ) is similar, or it follows from

- .