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 < ¹⁄₂r²sin(θ) < ¹⁄₂r²θ < ¹⁄₂r²tan(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 ${\displaystyle \theta \rightarrow 0,{\frac {\sin(\theta )}{\theta }}\rightarrow 1}$ and ${\displaystyle {\frac {\tan(\theta )}{\theta }}\rightarrow 1}$.

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

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

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,

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

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

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

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

Proof: For any h,

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

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

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

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

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