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9.1 Properties of Sine and Cosine

9.1   Definition ($W(t).$) We define a function $W\colon\mbox{{\bf R}}\to\mbox{{\bf R}}^2$ as follows.

If $t\geq 0$, then $W(t)$ is the point on the unit circle such that the length of the arc joining $(1,0)$ to $W(t)$ (measured in the counterclockwise direction) is equal to $t$. (There is an optical illusion in the figure. The length of segment $[0,t]$ is equal to the length of arc $W(0)W(t)$.)

Thus to find $W(t)$, you should start at $(1,0)$ and move along the circle in a counterclockwise direction until you've traveled a distance $t$. Since the circumference of the circle is $2\pi$, we see that $W(2\pi)$ $=W(4\pi)$ $=W(0)$ $=(1,0)$. (Here we assume Archimedes' result that the area of a circle is half the circumference times the radius.) If $t<0$, we define
W(t)=H(W(-t)) \mbox{ for } t<0
\end{displaymath} (9.2)

where $H$ is the reflection about the horizontal axis. Thus if $t<0$, then $W(t)$ is the point obtained by starting at $(1,0)$ and moving $\vert t\vert$ along the unit circle in the clockwise direction.

Remark: The definition of $W$ depends on several ideas that we have not defined or stated assumptions about, e.g., length of an arc and counterclockwise direction. I believe that the amount of work it takes to formalize these ideas at this point is not worth the effort, so I hope your geometrical intuition will carry you through this chapter. (In this chapter we will assume quite a bit of Euclidean geometry, and a few properties of area that do not follow from our assumptions stated in chapter 5.)

A more self contained treatment of the trigonometric functions can be found in [44, chapter 15], but the treatment given there uses ideas that we will consider later, (e.g. derivatives, inverse functions, the intermediate value theorem, and the fundamental theorem of calculus) in order to define the trigonometric functions.

We have the following values for $W$:


$\displaystyle W(0)$ $\textstyle =$ $\displaystyle (1,0)$ (9.3)
$\displaystyle W\Big({\pi\over 2}\Big)$ $\textstyle =$ $\displaystyle (0,1)$ (9.4)
$\displaystyle W(\pi)$ $\textstyle =$ $\displaystyle (-1,0)$ (9.5)
$\displaystyle W\Big({{3\pi}\over 2}\Big)$ $\textstyle =$ $\displaystyle (0,-1)$ (9.6)
$\displaystyle W(2\pi)$ $\textstyle =$ $\displaystyle (1,0)=W(0).$ (9.7)

In general

W(t+2\pi k)=W(t) \mbox{ for all } t\in\mbox{{\bf R}}\mbox{ and all }
k\in\mbox{{\bf Z}}.
\end{displaymath} (9.8)

9.9   Definition (Sine and cosine.) In terms of coordinates, we write

\begin{displaymath}W(t)=\Big(\cos (t),\sin (t)\Big).\end{displaymath}

(We read ``$\cos (t)$" as `` cosine of $t$", and we read ``$\sin
(t)$" as `` sine of $t$".)

Since $W(t)$ is on the unit circle, we have

\begin{displaymath}\sin^2 (t)+\cos^2 (t)=1 \mbox{ for all } t\in\mbox{{\bf R}},\end{displaymath}


\begin{displaymath}-1\leq\sin t\leq 1,\quad -1\leq\cos t\leq 1 \mbox{ for all } t\in\mbox{{\bf R}}.\end{displaymath}

The equations (9.3) - (9.8) show that

\cos (0)=1, & \sin(0)=0,\\
\cos\Big({\pi\o... 2}\Big)=0, & \sin\Big({{3\pi}\over 2}\Big) = -1,


\cos (t+2\pi k)=\cos t & \mbox{ for all } t\in\mbox{{\bf R}}\m...
...for all } t\in\mbox{{\bf R}}\mbox{ and all } k\in\mbox{{\bf Z}}.

In equation (9.2) we defined

\begin{displaymath}W(t)=H(W(-t)) \mbox{ for } t<0.\end{displaymath}

Thus for $t<0$,

\begin{displaymath}W(-t) = H(H(W(-t))) = H(W(t)) = H(W(-(-t))),\end{displaymath}

and it follows that

\begin{displaymath}W(t)= H(W(-t)) \mbox{ for all } t\in\mbox{{\bf R}}.\end{displaymath}

In terms of components

\Big(\cos (-t),\sin (-t)\Big) &=& W(-t)=H(W(t))=H(\cos (t),\si...
...& \Big( \cos (t),-\sin (t)\Big)
\index{trigonometric identities}

and consequently

\begin{displaymath}\cos (-t)=\cos (t) \mbox{ and } \sin (-t)=-\sin (t) \mbox{ for all } t\in\mbox{{\bf R}}.\end{displaymath}

Let $s,t$ be arbitrary real numbers. Then there exist integers $k$ and $l$ such that $s+2\pi k\in [0,2\pi)$ and $t+2\pi l\in [0,2\pi)$. Let

\begin{displaymath}s^\prime =s+2\pi k \mbox{ and } t^\prime =t+2\pi l.\end{displaymath}

Then $s^\prime -t^\prime =(s-t)+2\pi (k-l)$, so

\begin{displaymath}W(s)=W(s^\prime),\; W(t)=W(t^\prime),\; W(s-t)=W(s^\prime -t^\prime ).\end{displaymath}

Suppose $t^\prime\leq s^\prime$ (see figure). Then the length of the arc joining $W(s^\prime )$ to $W(t^\prime )$ is $s^\prime -t^\prime$ which is the same as the length of the arc joining $(1,0)$ to $W(s^\prime -t^\prime )$. Since equal arcs in a circle subtend equal chords, we have

\begin{displaymath}\mbox{dist}\Big(W(s^\prime ),W(t^\prime)\Big)=\mbox{dist}\Big(

and hence
\end{displaymath} (9.10)

You can verify that this same relation holds when $s^\prime <t^\prime$.

9.11   Theorem (Addition laws for sine and cosine.) For all real numbers $s$ and $t$,
$\displaystyle \cos(s+t)$ $\textstyle =$ $\displaystyle \cos (s)\cos (t)-\sin (s)\sin (t)$ (9.12)
$\displaystyle \cos (s-t)$ $\textstyle =$ $\displaystyle \cos (s)\cos (t)+\sin (s)\sin (t)$ (9.13)
$\displaystyle \sin (s+t)$ $\textstyle =$ $\displaystyle \sin (s)\cos (t)+\cos (s)\sin (t)$ (9.14)
$\displaystyle \sin (s-t)$ $\textstyle =$ $\displaystyle \sin (s) \cos (t)-\cos (s)\sin (t).$ (9.15)

Proof: From (9.10) we know

\begin{displaymath}\mbox{dist}\Big( W(s),W(t)\Big)=\mbox{dist}\Big( W(s-t),(1,0)\Big),\end{displaymath}


\begin{displaymath}\mbox{dist}\Big( (\cos (s),\sin (s)),(\cos (t),\sin


\begin{displaymath}\Big(\cos (s)-\cos (t)\Big)^2+\Big(\sin (s)-\sin (t)\Big)^2=\Big(\cos
(s-t)-1\Big)^2+\Big(\sin (s-t)\Big)^2.\end{displaymath}

By expanding the squares and using the fact that $\sin ^2(u)+\cos^2(u)=1$ for all $u$, we conclude that
\cos (s)\cos (t)+\sin (s)\sin (t)=\cos (s-t).
\end{displaymath} (9.16)

This is equation (9.13). To get equation (9.12) replace $t$ by $-t$ in (9.16). If we take $\displaystyle {s={\pi\over 2}}$ in equation (9.16) we get

\begin{displaymath}\cos\Big({\pi\over 2}\Big)\cos (t)+\sin\Big( {\pi\over 2}\Big) \sin
(t)=\cos\Big({\pi\over 2}-t\Big)\end{displaymath}


\begin{displaymath}\sin (t)=\cos \Big( {\pi\over 2}-t\Big) \mbox{ for all } t\in\mbox{{\bf R}}.\end{displaymath}

If we replace $t$ by $\displaystyle {\Big({\pi\over 2}-t\Big)}$ in this equation we get

\begin{displaymath}\sin\Big({\pi\over 2}-t\Big)=\cos\Big( {\pi\over 2}-({\pi\over 2}-t)\Big)=\cos
t \mbox{ for all } t\in\mbox{{\bf R}}.\end{displaymath}

Now in equation (9.16) replace $s$ by $\displaystyle {{\pi\over 2}-s}$ and get

\begin{displaymath}\cos\Big({\pi\over 2}-s\Big)\cos (t)+\sin \Big({\pi\over


\begin{displaymath}\sin s\cos t +\cos s\sin t=\sin (s+t),\end{displaymath}

which is equation (9.14). Finally replace $t$ by $-t$ in this last equation to get (9.15). $\diamondsuit$

In the process of proving the last theorem we proved the following:

9.17   Theorem (Reflection law for sin and cos.) For all $x \in \mbox{{\bf R}}$,
\cos(x) = \sin({\pi\over 2} - x) \mbox{ and }\sin(x) = \cos({\pi\over 2} - x).
\end{displaymath} (9.18)

9.19   Theorem (Double angle and half angle formulas.) For all $t\in\mbox{{\bf R}}$ we have

\sin(2t)=2\sin t\cos t,\\
\cos (2t)=\cos^2 ...
...e {\cos^2\Big({t\over 2}\Big)={{1+\cos t}\over 2}}.

9.20   Exercise. A Prove the four formulas stated in theorem 9.19.

9.21   Theorem (Products and differences of sin and cos.) For all $s,t$ in $\mbox{{\bf R}}$,
$\displaystyle \cos(s)\cos(t)$ $\textstyle =$ $\displaystyle {1\over 2}[\cos (s-t)+\cos (s+t)],$ (9.22)
$\displaystyle \cos(s)\sin(t)$ $\textstyle =$ $\displaystyle {1\over 2}[\sin (s+t)-\sin (s-t)],$ (9.23)
$\displaystyle \sin(s)\sin(t)$ $\textstyle =$ $\displaystyle {1\over 2}[\cos (s-t)-\cos (s+t)]
\index{trigonometric identities},$ (9.24)
$\displaystyle \cos(s)-\cos(t)$ $\textstyle =$ $\displaystyle -2\sin\Big({{s+t}\over 2}\Big)\sin\Big({{s-t}\over
2}\Big),$ (9.25)
$\displaystyle \sin(s)-\sin(t)$ $\textstyle =$ $\displaystyle 2\cos\Big( {{s+t}\over 2}\Big)\sin\Big({{s-t}\over
2}\Big).$ (9.26)

Proof: We have

\begin{displaymath}\cos (s+t)=\cos (s)\cos (t)-\sin (s)\sin (t)\end{displaymath}


\begin{displaymath}\cos (s-t)=\cos (s)\cos (t)+\sin (s)\sin (t).\end{displaymath}

By adding these equations, we get (9.22). By subtracting the first from the second, we get (9.24).

In equation (9.24) replace $s$ by $\displaystyle { {{s+t}\over 2}}$ and replace $t$ by $\displaystyle {
{{t-s}\over 2}}$ to get

\begin{displaymath}\sin\Big( {{s+t}\over 2}\Big)\sin\Big({{t-s}\over 2}\Big)
={1... 2}\Big)
- \cos\Big( {{s+t}\over 2}+{{t-s}\over 2}\Big)\Big]\end{displaymath}


\begin{displaymath}-\sin\Big( {{s+t}\over 2}\Big)\sin\Big({{s-t}\over 2}\Big)={1\over 2}[\cos
(s)-\cos (t)].\end{displaymath}

This yields equation (9.25).

9.27   Exercise. Prove equations (9.23) and (9.26).

From the geometrical description of sine and cosine, it follows that as $t$ increases for $0$ to $\displaystyle {{\pi\over 2}}$, $\sin
(t)$ increases from $0$ to $1$ and $\cos (t)$ decreases from $1$ to $0$. The identities

\begin{displaymath}\sin\Big( {\pi\over 2}-t\Big)=\cos (t) \mbox{ and } \cos\Big({\pi\over

indicate that a reflection about the vertical line through $\displaystyle {x={\pi\over
4}}$ carries the graph of sin onto the graph of cos, and vice versa.

The condition $\cos (-x)=\cos x$ indicates that the reflection about the vertical axis carries the graph of $\cos$ to itself.

The relation $\sin (-x)=-\sin(x)$ shows that

(x,y)\in\mbox{\rm graph}(\sin) &\mbox{$\Longrightarrow$}& y=\s...
&\mbox{$\Longrightarrow$}& R_\pi (x,y)\in\mbox{\rm graph}(\sin)

i.e., the graph of $\sin$ is carried onto itself by a rotation through $\pi$ about the origin.

We have

\begin{displaymath}\sin\Big({\pi\over 4}\Big)=\cos\Big({\pi\over 2}-{\pi\over

and $\displaystyle {1=\sin^2\Big({\pi\over 4}\Big)+\cos^2\Big({\pi\over
4}\Big) =2\cos^2\Big({\pi\over 4}\Big)}$, so $\displaystyle {\cos^2\Big({\pi\over
2}}$ and
$\displaystyle {\sin{\pi\over 4}=\cos{\pi\over 4}={{\sqrt 2}\over
2}=.707}$ (approximately).

With this information we can make a reasonable sketch of the graph of $\sin$ and $\cos$ (see the figure above)

9.28   Exercise. Show that

\begin{displaymath}\cos(3x) = 4 \cos^3(x) - 3\cos(x) \mbox{ for all }x \in \mbox{{\bf R}}.\end{displaymath}

9.29   Exercise. A Complete the following table of sines and cosines:

\begin{displaymath}\begin{array}{\vert l\vert l\vert l\vert l\vert l\vert l\vert...
... &{{\sqrt 2}\over 2} & &0 & & & &-1  [2ex] \hline

\begin{displaymath}\begin{array}{\vert l\vert l\vert l\vert l\vert l\vert l\vert...
\cos &-1 & & & & 0 & & & & 1  [2ex] \hline

\begin{displaymath}{{\sqrt 2}\over 2}=.707\end{displaymath}

Include an explanation for how you found $\displaystyle {\sin{\pi\over 6}}$ and $\displaystyle {\cos{\pi\over
6}}$ (or $\displaystyle {\sin{\pi\over 3}}$ and $\displaystyle {\cos{\pi\over 3}}$). For the remaining values you do not need to include an explanation.

Most of the material from this section was discussed by Claudius Ptolemy (fl. 127-151 AD). The functions considered by Ptolemy were not the sine and cosine, but the chord, where the chord of an arc $\alpha$ is the length of the segment joining its endpoints.


\mbox{chord }(\alpha )=2\sin ({\alpha\over 2}).
\end{displaymath} (9.30)

Ptolemy's chords are functions of arcs (measured in degrees), not of numbers. Ptolemy's addition law for $\sin$ was (roughly)

\begin{displaymath}D \cdot\mbox { chord} (\beta -\alpha )= \mbox{chord}(\beta
...- \alpha)-\mbox{chord}(180^\circ -\beta)
\mbox{chord}(\alpha ),\end{displaymath}

where $D$ is the diameter of the circle, and $0^\circ < \alpha < \beta
< 180^\circ.$ Ptolemy produced tables equivalent to tables of $\sin (\alpha)$ for $\displaystyle {\left({1\over
4}\right)^\circ\leq\alpha\leq 90^\circ}$ in intervals of $\displaystyle {\left({1\over 4}
\right)^\circ}$. All calculations were made to 3 sexagesimal (base $60$) places.

The etymology of the word sine is rather curious[42, pp 615-616]. The function we call sine was first given a name by Aryabhata near the start of the sixth century AD. The name meant ``half chord'' and was later shortened to jya meaning ``chord''. The Hindu word was translated into Arabic as jîba, which was a meaningless word phonetically derived from jya, but (because the vowels in Arabic were not written) was written the same as jaib, which means bosom. When the Arabic was translated into Latin it became sinus. (Jaib means bosom, bay, or breast: sinus means bosom, bay or the fold of a toga around the breast.) The English word sine is derived from sinus phonetically.

9.31   Entertainment (Calculation of sines.) Design a computer program that will take as input a number $x$ between $0$ and $.5$, and will calculate $\sin (\pi x)$. (I choose $\sin (\pi x)$ instead of $\sin (x)$ so that you do not need to know the value of $\pi$ to do this.)

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Next: 9.2 Calculation of Up: 9. Trigonometric Functions Previous: 9. Trigonometric Functions   Index
Ray Mayer 2007-09-07