Trigonometry/Some preliminary results

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

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

Proof: Consider a circle, centre ${\displaystyle O}$, radius ${\displaystyle r}$, and choose two points ${\displaystyle A}$ and ${\displaystyle B}$ on the circumference such that ${\displaystyle \angle AOB}$ is less than a right angle. Draw a tangent to the circle at ${\displaystyle B}$, and let ${\displaystyle {\overrightarrow {OA}}}$ produced intersect it at ${\displaystyle C}$. Clearly

${\displaystyle 0<\operatorname {area} \left(\triangle OAB\right)<\operatorname {area} \left({\text{⌔}}OAB\right)<\operatorname {area} \left(\triangle OBC\right)}$

i.e.

${\displaystyle 0<{\frac {1}{2}}r^{2}\sin(\theta )<{\frac {1}{2}}r^{2}\theta <{\frac {1}{2}}r^{2}\tan(\theta )}$

and the result follows.

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

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

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

Proof: Dividing the result of the previous theorem by ${\displaystyle \sin(\theta )}$ and taking reciprocals,

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

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

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

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

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

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

Proof:

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

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

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

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

Proof:

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

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

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

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

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

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

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

Proof: For any ${\displaystyle h}$,

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

since ${\displaystyle \left\vert \cos(\theta )\right\vert }$ cannot exceed ${\displaystyle 1}$ and ${\displaystyle \left\vert \sin(\theta )\right\vert }$ cannot exceed ${\displaystyle \left\vert x\right\vert }$. Thus, as

${\displaystyle h\rightarrow 0,\,\sin(\theta +h)\rightarrow \sin(\theta )}$,

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

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