# Trigonometry/Some preliminary results

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 $\theta \rightarrow 0, \frac{\sin(\theta)}{\theta} \rightarrow 1$ and $\frac{\tan(\theta)}{\theta} \rightarrow 1$.

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

$1 > \frac{\sin(\theta)}{\theta} > \cos(\theta)$.

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,

$\frac{1}{\cos(\theta)} > \frac{\tan(\theta)}{\theta} > 1$.

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

Theorem: If θ is as before then $\cos(\theta) > 1-\frac{\theta^2}{2}$.

Proof:

$\cos(\theta) = 1-2\sin^2 \left( \frac{\theta}{2} \right)$
$\sin \left( \frac{\theta}{2} \right) < \frac{\theta}{2} \text{ so}$
$\cos(\theta) > 1-2\left( \frac{\theta}{2} \right)^2 = 1-\frac{\theta^2}{2}$.

Theorem: If θ is as before then $\sin(\theta) > \theta-\frac{\theta^3}{4}$.

Proof:

$\sin (\theta) = 2\sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right) = 2\tan \left( \frac{\theta}{2} \right) \cos^2 \left( \frac{\theta}{2} \right)$.
$\tan \left( \frac{\theta}{2} \right) > \frac{\theta}{2} \text{ so}$
$\sin (\theta) > 2 \left( \frac{\theta}{2} \right) \cos^2 \left( \frac{\theta}{2} \right) = \theta \left( 1- \sin^2 \left( \frac{\theta}{2} \right) \right)$.
$\sin \left( \frac{\theta}{2} \right) < \frac{\theta}{2} \text{ so}$
$1-\sin^2 \left( \frac{\theta}{2} \right) > 1-\left( \frac{\theta}{2} \right)^2 \text{ so}$
$\sin(\theta) < \theta \left( 1 - \left( \frac{\theta}{2} \right)^2 \right) = \theta - \frac{\theta^3}{4}$.

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

Proof: For any h,

$|\sin(\theta+h) - \sin(\theta)| = 2|\cos \left(\theta+\frac{h}{2} \right)||\sin \left(\frac{h}{2} \right)| < h$,

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

$h \rightarrow 0, \, \sin(\theta+h) \rightarrow \sin(\theta)$,

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

$\cos (\theta) = \sin \left(\frac{\pi}{2} - \theta \right)$.