We prove some results that are needed in the application of calculus to trigonometry.
Reference Image for Proof
Theorem: If
θ
{\displaystyle \theta }
; is a positive angle, less than a right angle (expressed in radians), then
0
<
sin
(
θ
)
<
θ
<
tan
(
θ
)
{\displaystyle 0<\sin(\theta )<\theta <\tan(\theta )}
.
Proof: Consider a circle, centre
O
{\displaystyle O}
, radius
r
{\displaystyle r}
, and choose two points
A
{\displaystyle A}
and
B
{\displaystyle B}
on the circumference such that
∠
A
O
B
{\displaystyle \angle AOB}
is less than a right angle. Draw a tangent to the circle at
B
{\displaystyle B}
, and let
O
A
→
{\displaystyle {\overrightarrow {OA}}}
produced intersect it at
C
{\displaystyle C}
. Clearly
0
<
area
(
△
O
A
B
)
<
area
(
⌔
O
A
B
)
<
area
(
△
O
B
C
)
{\displaystyle 0<\operatorname {area} \left(\triangle OAB\right)<\operatorname {area} \left({\text{⌔}}OAB\right)<\operatorname {area} \left(\triangle OBC\right)}
i.e.
0
<
1
2
r
2
sin
(
θ
)
<
1
2
r
2
θ
<
1
2
r
2
tan
(
θ
)
{\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
θ
{\displaystyle \theta }
is a negative angle, more than minus a right angle (expressed in radians), then
0
>
sin
(
θ
)
>
θ
>
tan
(
θ
)
{\displaystyle 0>\sin(\theta )>\theta >\tan(\theta )}
. [This follows from
sin
(
−
θ
)
=
−
sin
(
θ
)
{\displaystyle \sin(-\theta )=-\sin(\theta )}
and
tan
(
−
θ
)
=
−
tan
(
θ
)
{\displaystyle \tan(-\theta )=-\tan(\theta )}
.]
Corollary: If
θ
{\displaystyle \theta }
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
|
{\displaystyle 0<\left\vert \sin \theta \right\vert <\left\vert \theta \right\vert <\left\vert \tan \theta \right\vert }
.
Theorem: As
θ
→
0
,
sin
(
θ
)
θ
→
1
{\displaystyle \theta \rightarrow 0,{\frac {\sin(\theta )}{\theta }}\rightarrow 1}
and
tan
(
θ
)
θ
→
1
{\displaystyle {\frac {\tan(\theta )}{\theta }}\rightarrow 1}
.
Proof: Dividing the result of the previous theorem by
sin
(
θ
)
{\displaystyle \sin(\theta )}
and taking reciprocals,
1
>
sin
(
θ
)
θ
>
cos
(
θ
)
{\displaystyle 1>{\frac {\sin(\theta )}{\theta }}>\cos(\theta )}
.
But
cos
(
θ
)
{\displaystyle \cos(\theta )}
tends to
1
{\displaystyle 1}
as
θ
{\displaystyle \theta }
tends to
0
{\displaystyle 0}
, so the first part follows.
Dividing the result of the previous theorem by
tan
(
θ
)
{\displaystyle \tan(\theta )}
and taking reciprocals,
1
cos
(
θ
)
>
tan
(
θ
)
θ
>
1
{\displaystyle {\frac {1}{\cos(\theta )}}>{\frac {\tan(\theta )}{\theta }}>1}
.
Again,
cos
(
θ
)
{\displaystyle \cos(\theta )}
tends to
1
{\displaystyle 1}
as
θ
{\displaystyle \theta }
tends to
0
{\displaystyle 0}
, so the second part follows.
Theorem: If
θ
{\displaystyle \theta }
is as before, then
cos
(
θ
)
>
1
−
θ
2
2
{\displaystyle \cos(\theta )>1-{\frac {\theta ^{2}}{2}}}
.
Proof:
cos
(
θ
)
=
1
−
2
sin
2
(
θ
2
)
{\displaystyle \cos(\theta )=1-2\sin ^{2}\left({\frac {\theta }{2}}\right)}
sin
(
θ
2
)
<
θ
2
so
{\displaystyle \sin \left({\frac {\theta }{2}}\right)<{\frac {\theta }{2}}{\text{ so}}}
cos
(
θ
)
>
1
−
2
(
θ
2
)
2
=
1
−
θ
2
2
{\displaystyle \cos(\theta )>1-2\left({\frac {\theta }{2}}\right)^{2}=1-{\frac {\theta ^{2}}{2}}}
.
Theorem: If
θ
{\displaystyle \theta }
is as before, then
sin
(
θ
)
>
θ
−
θ
3
4
{\displaystyle \sin(\theta )>\theta -{\frac {\theta ^{3}}{4}}}
.
Proof:
sin
(
θ
)
=
2
sin
(
θ
2
)
cos
(
θ
2
)
=
2
tan
(
θ
2
)
cos
2
(
θ
2
)
{\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)}
.
tan
(
θ
2
)
>
θ
2
so
{\displaystyle \tan \left({\frac {\theta }{2}}\right)>{\frac {\theta }{2}}{\text{ so}}}
sin
(
θ
)
>
2
(
θ
2
)
cos
2
(
θ
2
)
=
θ
(
1
−
sin
2
(
θ
2
)
)
{\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)}
.
sin
(
θ
2
)
<
θ
2
so
{\displaystyle \sin \left({\frac {\theta }{2}}\right)<{\frac {\theta }{2}}{\text{ so}}}
1
−
sin
2
(
θ
2
)
>
1
−
(
θ
2
)
2
so
{\displaystyle 1-\sin ^{2}\left({\frac {\theta }{2}}\right)>1-\left({\frac {\theta }{2}}\right)^{2}{\text{ so}}}
sin
(
θ
)
<
θ
(
1
−
(
θ
2
)
2
)
=
θ
−
θ
3
4
{\displaystyle \sin(\theta )<\theta \left(1-\left({\frac {\theta }{2}}\right)^{2}\right)=\theta -{\frac {\theta ^{3}}{4}}}
.
Theorem:
sin
(
θ
)
{\displaystyle \sin(\theta )}
and
cos
(
θ
)
{\displaystyle \cos(\theta )}
are continuous functions.
Proof: For any
h
{\displaystyle h}
,
|
sin
(
θ
+
h
)
−
sin
(
θ
)
|
=
2
|
cos
(
θ
+
h
2
)
|
|
sin
(
h
2
)
|
<
h
{\displaystyle |\sin(\theta +h)-\sin(\theta )|=2|\cos \left(\theta +{\frac {h}{2}}\right)||\sin \left({\frac {h}{2}}\right)|<h}
,
since
|
cos
(
θ
)
|
{\displaystyle \left\vert \cos(\theta )\right\vert }
cannot exceed
1
{\displaystyle 1}
and
|
sin
(
θ
)
|
{\displaystyle \left\vert \sin(\theta )\right\vert }
cannot exceed
|
x
|
{\displaystyle \left\vert x\right\vert }
. Thus, as
h
→
0
,
sin
(
θ
+
h
)
→
sin
(
θ
)
{\displaystyle h\rightarrow 0,\,\sin(\theta +h)\rightarrow \sin(\theta )}
,
proving continuity. The proof for cos(θ) is similar, or it follows from
cos
(
θ
)
=
sin
(
π
2
−
θ
)
{\displaystyle \cos(\theta )=\sin \left({\frac {\pi }{2}}-\theta \right)}
.