Trigonometry/Some preliminary results

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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 < 12r2sin(θ) < 12r2θ < 12r2tan(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

.