# 8.3: The Sine and Cosine Functions

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We begin by defining functions $s:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$

and

$c:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$

by

$s(x)=\left\{\begin{array}{ll}{\frac{\tan (x)}{\sqrt{1+\tan ^{2}(x)}},} & {\text { if } x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)}, \\ {1,} & {\text { if } x=\frac{\pi}{2}}\end{array}\right.$

and

$c(x)=\left\{\begin{array}{ll}{\frac{1}{\sqrt{1+\tan ^{2}(x)}},} & {\text { if } x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)}, \\ {0,} & {\text { if } x=\frac{\pi}{2}}.\end{array}\right.$

Note that

$\lim _{x \rightarrow \frac{\pi}{2}-\pi^{-}} s(x)=\lim _{y \rightarrow+\infty} \frac{y}{\sqrt{1+y^{2}}}=\lim _{y \rightarrow+\infty} \frac{1}{\sqrt{1+\frac{1}{y^{2}}}}=1$

and

$\lim _{x \rightarrow \frac{\pi}{2}-} c(x)=\lim _{y \rightarrow+\infty} \frac{1}{\sqrt{1+y^{2}}}=\lim _{y \rightarrow+\infty} \frac{\frac{1}{y}}{\sqrt{1+\frac{1}{y^{2}}}}=0,$

which shows that both $$s$$ and $$c$$ are continuous functions.

Next, we extend the definitions of $$s$$ and $$c$$ to functions

$S:\left(-\frac{\pi}{2}, \frac{3 \pi}{2}\right] \rightarrow \mathbb{R}$

and

$C:\left(-\frac{\pi}{2}, \frac{3 \pi}{2}\right] \rightarrow \mathbb{R}$

by defining

$S(x)=\left\{\begin{array}{ll}{s(x),} & {\text { if } x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right]}, \\ {-s(x-\pi),} & {\text { if } x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right]}\end{array}\right.$

and

$C(x)=\left\{\begin{array}{ll}{c(x),} & {\text { if } x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right]}, \\ {-c(x-\pi),} & {\text { if } x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right]}.\end{array}\right.$

Note that

$\lim _{x \rightarrow \frac{\pi}{2}+} S(x)=\lim _{x \rightarrow-\frac{\pi}{2}+}-s(x)=-\lim _{y \rightarrow-\infty} \frac{y}{\sqrt{1+y^{2}}}=-\lim _{y \rightarrow-\infty} \frac{1}{-\sqrt{1+\frac{1}{y^{2}}}}=1$

and

$\lim _{x \rightarrow \frac{\pi}{2}+} C(x)=\lim _{x \rightarrow-\frac{\pi}{2}+}-c(x)=-\lim _{y \rightarrow-\infty} \frac{1}{\sqrt{1+y^{2}}}=-\lim _{y \rightarrow-\infty} \frac{\frac{1}{y}}{-\sqrt{1+\frac{1}{y^{2}}}}=0,$

which shows that both $$S$$ and $$C$$ are continuous at $$\frac{\pi}{2} .$$ Thus both $$S$$ and $$C$$ are continuous.

Finally, for any $$x \in \mathbb{R},$$ let

$g(x)=\sup \left\{n: n \in \mathbb{Z},-\frac{\pi}{2}+2 n \pi<x\right\}$

and define

$\sin (x)=S(x-2 \pi g(x))$

and

$\cos (x)=C(x-2 \pi g(x)).$

##### Definition

With the notation as above, for any $$x \in \mathbb{R},$$ we call $$\sin (x)$$ and $$\cos (x)$$ the sine and cosine of $$x,$$ respectively.

##### Proposition $$\PageIndex{1}$$

The sine and cosine functions are continuous on $$\mathbb{R}$$.

Proof

From the definitions, it is sufficient to verify continuity at $$\frac{3 \pi}{2} .$$ Now

$\lim _{x \rightarrow \frac{3 \pi}{2}-} \sin (x)=\lim _{x \rightarrow \frac{3 \pi}{2}-} S(x)=S\left(\frac{3 \pi}{2}\right)=-s\left(\frac{\pi}{2}\right)=-1$

and

\begin{aligned} \lim _{x \rightarrow \frac{3 \pi}{2}+} \sin (x) &=\lim _{x \rightarrow \frac{3\pi}{2}+} S(x-2 \pi) \\ &=\lim _{x \rightarrow-\frac{\pi}{2}+} s(x) \\ &=\lim _{y \rightarrow-\infty} \frac{y}{\sqrt{1+y^{2}}} \\ &=\lim _{y \rightarrow-\infty} \frac{1}{-\sqrt{1+\frac{1}{y^{2}}}} \\ &=-1, \end{aligned}

and so sine is continuous at $$\frac{3 \pi}{2}$$. Similarly,

$\lim _{x \rightarrow \frac{3 \pi}{2}-} \cos (x)=\lim _{x \rightarrow \frac{3 \pi}{2}-} C(x)=C\left(\frac{3 \pi}{2}\right)=-c\left(\frac{\pi}{2}\right)=0$

and

\begin{aligned} \lim _{x \rightarrow \frac{3 \pi}{2}+} \cos (x) &=\lim _{x \rightarrow \frac{3 \pi}{2}+} C(x-2 \pi) \\ &=\lim _{x \rightarrow-\frac{\pi}{2}+} c(x) \\ &=\lim _{y \rightarrow-\infty} \frac{1}{\sqrt{1+y^{2}}} \\ &=\lim _{y \rightarrow-\infty} \frac{\frac{1}{y}}{-\sqrt{1+\frac{1}{y^{2}}}} \\ &=0, \end{aligned}

and so cosine is continuous at $$\frac{3 \pi}{2}$$. $$\quad$$ Q.E.D.

## 8.3.1 Properties of sine and cosine

##### Proposition $$\PageIndex{2}$$

The sine and cosine functions are periodic with period $$2\pi$$.

Proof

The result follows immediately from the definitions. $$\quad$$ Q.E.D.

##### Proposition $$\PageIndex{3}$$

For any $$x \in \mathbb{R}, \sin (-x)=-\sin (x)$$ and $$\cos (-x)=\cos (x) .$$

Proof

The result follows immediately from the definitions. $$\quad$$ Q.E.D.

##### Proposition $$\PageIndex{4}$$

For any $$x \in \mathbb{R}, \sin ^{2}(x)+\cos ^{2}(x)=1$$.

Proof

The result follows immediately from the definition of $$s$$ and $$c .$$ $$\quad$$ Q.E.D.

##### Proposition $$\PageIndex{5}$$

The range of both the sine and cosine functions is $$[-1,1] .$$

Proof

The result follows immediately from the definitions along with the facts that

$\sqrt{1+y^{2}} \geq \sqrt{y^{2}}=|y|$

and

$\sqrt{1+y^{2}} \geq 1$

for any $$y \in \mathbb{R} . \quad$$ Q.E.D

##### Proposition $$\PageIndex{6}$$

For any $$x$$ in the domain of the tangent function,

$\tan (x)=\frac{\sin (x)}{\cos (x)}.$

Proof

The result follows immediately from the definitions. $$\quad$$ Q.E.D.

##### Proposition $$\PageIndex{7}$$

For any $$x$$ in the domain of the tangent function,

$\sin ^{2}(x)=\frac{\tan ^{2}(x)}{1+\tan ^{2}(x)}$

and

$\cos ^{2}(x)=\frac{1}{1+\tan ^{2}(x)}.$

Proof

The result follows immediately from the definitions. $$\quad$$ Q.E.D.

##### Proposition $$\PageIndex{8}$$

For any $$x, y \in \mathbb{R}$$,

$\cos (x+y)=\cos (x) \cos (y)-\sin (x) \sin (y).$

Proof

First suppose $$x, y,$$ and $$x+y$$ are in the domain of the tangent function.

Then

\begin{aligned} \cos^{2}(x+y) &= \frac{1}{1+\tan^{2}(x+y)} \\ &= \frac{1}{1+\left(\frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)} \right)^{2}} \\ &= \frac{(1 - \tan(x)\tan(y))^{2}}{1 - \tan(x)\tan(y))^{2} + (\tan(x) + \tan(y))^{2}} \\ &= \frac{(1 - \tan(x)\tan(y))^{2}}{(1+\tan^{2}(x))(1+\tan^{2}(y))} \\ &= \left(\frac{1}{\sqrt{1+\tan^{2}(x)} \sqrt{1+\tan^{2}(y)}} - \frac{\tan(x)\tan(y)}{\sqrt{1+\tan^{2}(x)}\sqrt{1+\tan^{2}(y)}}\right)^{2} \\ &= (\cos(x)\cos(y) - \sin(x)\sin(y))^{2} . \end{aligned}

Hence

$\cos (x+y)=\pm(\cos (x) \cos (y)-\sin (x) \sin (y)).$

Consider a fixed value of $$x$$. Note that the positive sign must be chosen when $$y=0 .$$ Moreover, increasing $$y$$ by $$\pi$$ changes the sign on both sides, so the positive sign must be chosen when $$y$$ is any multiple of $$\pi$$. Since sine and cosine are continuous functions, the choice of sign could change only at points at which both sides are $$0,$$ but these points are separated by a distance of $$\pi,$$ so we must always choose the positive sign. Hence we have

$\cos (x+y)=\cos (x) \cos (y)-\sin (x) \sin (y)$

for all $$x, y \in \mathbb{R}$$ for which $$x, y,$$ and $$x+y$$ are in the domain of the tangent function. The identity for the other values of $$x$$ and $$y$$ now follows from the continuity of the sine and cosine functions. $$\quad$$ Q.E.D.

##### Proposition $$\PageIndex{9}$$

For any $$x, y \in \mathbb{R}$$,

$\sin (x+y)=\sin (x) \cos (y)+\sin (y) \cos (x).$

##### Exercise $$\PageIndex{1}$$

Prove the previous proposition.

##### Exercise $$\PageIndex{2}$$

Show that for any $$x \in \mathbb{R}$$,

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

and

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

##### Exercise $$\PageIndex{3}$$

Show that for any $$x \in \mathbb{R}$$,

$\sin (2 x)=2 \sin (x) \cos (x)$

and

$\cos (2 x)=\cos ^{2}(x)-\sin ^{2}(x).$

##### Exercise $$\PageIndex{4}$$

Show that for any $$x \in \mathbb{R}$$,

$\sin ^{2}(x)=\frac{1-\cos (2 x)}{2}$

and

$\cos ^{2}(x)=\frac{1+\cos (2 x)}{2}.$

##### Exercise $$\PageIndex{5}$$

Show that

$\sin \left(\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}},$

$\sin \left(\frac{\pi}{6}\right)=\cos \left(\frac{\pi}{3}\right)=\frac{1}{2},$

and

$\sin \left(\frac{\pi}{3}\right)=\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}.$

## 8.3.2 The calculus of the trigonometric functions

##### Proposition $$\PageIndex{10}$$

$$\lim _{x \rightarrow 0} \frac{\arctan (x)}{x}=1$$.

Proof

Using l'Hôpital's rule,

$\lim _{x \rightarrow 0} \frac{\arctan (x)}{x}=\lim _{x \rightarrow 0} \frac{\frac{1}{1+x^{2}}}{1}=1.$

Q.E.D

##### Proposition $$\PageIndex{11}$$

$$\lim _{x \rightarrow 0} \frac{\tan (x)}{x}=1$$.

Proof

Letting $$x=\arctan (u),$$ we have

$\lim _{x \rightarrow 0} \frac{\tan (x)}{x}=\lim _{u \rightarrow 0} \frac{u}{\arctan (u)}=1.$

Q.E.D

##### Proposition $$\PageIndex{12}$$

$$\lim _{x \rightarrow 0} \frac{\sin (x)}{x}=1$$.

Proof

We have

$\lim _{x \rightarrow 0} \frac{\sin (x)}{x}=\lim _{x \rightarrow 0} \frac{\tan (x)}{x} \cos (x)=1.$

##### Proposition $$\PageIndex{13}$$

$$\lim _{x \rightarrow 0} \frac{1-\cos (x)}{x}=0 .$$

Proof

We have

\begin{aligned} \lim _{x \rightarrow 0} \frac{1-\cos (x)}{x} &=\lim _{x \rightarrow 0}\left(\frac{1-\cos (x)}{x}\right)\left(\frac{1+\cos (x)}{1+\cos (x)}\right) \\ &=\lim _{x \rightarrow 0} \frac{1-\cos ^{2}(x)}{x(1+\cos (x))} \\ &=\lim _{x \rightarrow 0}\left(\frac{\sin (x)}{x}\right)\left(\frac{\sin (x)}{1+\cos (x)}\right) \\ &=(1)(0) \\ &=0. \end{aligned}

Q.E.D

##### Proposition $$\PageIndex{14}$$

If $$f(x)=\sin (x),$$ then $$f^{\prime}(x)=\cos (x) .$$

Proof

We have

\begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\sin (x) \cos (h)+\sin (h) \cos (x)-\sin (x)}{h} \\ &=\sin (x) \lim _{h \rightarrow 0} \frac{\cos (h)-1}{h}+\cos (x) \lim _{h \rightarrow 0} \frac{\sin (h)}{h} \\ &=\cos (x). \end{aligned}

Q.E.D

##### Proposition $$\PageIndex{15}$$

If $$f(x)=\cos (x),$$ then $$f^{\prime}(x)=-\sin (x)$$.

##### Exercise $$\PageIndex{6}$$

Prove the previous proposition.

##### Definition

For appropriate $$x \in \mathbb{R},$$ we call

$\cot (x)=\frac{\cos x}{\sin (x)},$

$\sec (x)=\frac{1}{\cos (x)},$

and

$\csc (x)=\frac{1}{\sin (x)}$

the cotangent, secant, and cosecant of $$x,$$ respectively.

##### Exercise $$\PageIndex{7}$$

If $$f(x)=\tan (x)$$ and $$g(x)=\cot (x),$$ show that

$f^{\prime}(x)=\sec ^{2}(x)$

and

$g^{\prime}(x)=-\csc ^{2}(x).$

##### Exercise $$\PageIndex{8}$$

If $$f(x)=\sec (x)$$ and $$g(x)=\csc (x),$$ show that

$f^{\prime}(x)=\sec (x) \tan (x),$

and

$g^{\prime}(x)=-\csc (x) \cot (x).$

##### Proposition $$\PageIndex{16}$$

2 $$\int_{-1}^{1} \sqrt{1-x^{2}} d x=\pi$$.

Proof

Let $$x=\sin (u)$$. Then as $$u$$ varies from $$-\frac{\pi}{2}$$ to $$\frac{\pi}{2}, x$$ varies from $$-1$$ to $$1 .$$ And, for these values, we have

$\sqrt{1-x^{2}}=\sqrt{1-\sin ^{2}(u)}=\sqrt{\cos ^{2}(u)}=|\cos (u)|=\cos (u).$

Hence

\begin{aligned} \int_{-1}^{1} \sqrt{1-x^{2}} d x &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2}(u) d u \\ &=\int_{-\frac{\pi}{2}}^{\frac{1}{2}} \frac{1+\cos (2 u)}{2} d u \\ &=\int_{-\frac{\pi}{2}}^{\frac{1}{2}} \frac{1}{2} d u+\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos (2 u) d u \\ &=\frac{\pi}{2}+\frac{1}{4}(\sin (\pi)-\sin (-\pi)) \\ &=\frac{\pi}{2}. \end{aligned}

Q.E.D.

##### Exercise $$\PageIndex{9}$$

Find the Taylor polynomial $$P_{9}$$ of order 9 for $$f(x)=\sin (x)$$ at $$0 .$$ Note that this is equal to the Taylor polynomial of order 10 for $$f$$ at $$0 .$$ Is $$P_{9}\left(\frac{1}{2}\right)$$ an overestimate or an underestimate for sin $$\left(\frac{1}{2}\right) ?$$ Find an upper bound for the error in this approximation.

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