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 $\angle AOB$ 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(θ) is similar, or it follows from

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