Trigonometry/Some preliminary results

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



Theorem: If $$\theta$$; is a positive angle, less than a right angle (expressed in radians), then $$0 < \sin(\theta) < \theta < \tan(\theta)$$.

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 $$\overrightarrow{OA}$$ produced intersect it at $$C$$. Clearly
 * $$0 < \operatorname{area}\left(\triangle OAB\right) < \operatorname{area}\left(\text{⌔} OAB\right) < \operatorname{area}\left(\triangle OBC\right)$$

i.e.
 * $$0<\frac{1}{2}r^{2}\sin(\theta)<\frac{1}{2}r^{2}\theta<\frac{1}{2}r^{2}\tan(\theta)$$

and the result follows.

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

Corollary: If $$\theta$$ is a non-zero angle, less than a right angle but more than minus a right angle (expressed in radians), then $$0<\left\vert\sin\theta\right\vert<\left\vert\theta\right\vert<\left\vert\tan\theta\right\vert$$.

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 $$\sin(\theta)$$ and taking reciprocals,


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

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

Dividing the result of the previous theorem by $$\tan(\theta)$$ and taking reciprocals,


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

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

Theorem: If $$\theta$$ 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 $$\theta$$ 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: $$\sin(\theta)$$ and $$\cos(\theta)$$ 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 $$\left\vert\cos(\theta)\right\vert$$ cannot exceed $$1$$ and $$\left\vert\sin(\theta)\right\vert$$ cannot exceed $$\left\vert x\right\vert$$. Thus, as


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

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


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