Trigonometry/Some preliminary results

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

Reference Image for Proof

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 $\mathrm{\angle }AOB$ is less than a right angle. Draw a tangent to the circle at $B$, and let $\overrightarrow{OA}$ produced intersect it at $C$. Clearly

$0<\mathrm{area}\left(\mathrm{△}OAB\right)<\mathrm{area}\left(\text{⌔}OAB\right)<\mathrm{area}\left(\mathrm{△}OBC\right)$

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 $0<|\sin \theta |<\left|\theta \right|<|\tan \theta |$.

Theorem: As $\theta \to 0,\frac{\sin (\theta )}{\theta }\to 1$ and $\frac{\tan (\theta )}{\theta }\to 1$.

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:

$\sin (\theta )=2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)=2\tan \left(\frac{\theta }{2}\right){\cos }^2\left(\frac{\theta }{2}\right)$.

$\tan \left(\frac{\theta }{2}\right)>\frac{\theta }{2}\text{ so}$

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

$\sin \left(\frac{\theta }{2}\right)<\frac{\theta }{2}\text{ so}$

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

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

Theorem: $\sin (\theta )$ and $\cos (\theta )$ are continuous functions.

Proof: For any $h$,

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

since $|\cos (\theta )|$ cannot exceed $1$ and $|\sin (\theta )|$ cannot exceed $\left|x\right|$. Thus, as

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

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

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