8.3: The Sine and Cosine Functions
- Page ID
- 22687
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
\[\cos (x)=C(x-2 \pi g(x)).\]
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.
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
The sine and cosine functions are periodic with period \(2\pi\).
- Proof
-
The result follows immediately from the definitions. \(\quad\) Q.E.D.
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.
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.
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
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.
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.
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.
For any \(x, y \in \mathbb{R}\),
\[\sin (x+y)=\sin (x) \cos (y)+\sin (y) \cos (x).\]
Prove the previous proposition.
Show that for any \(x \in \mathbb{R}\),
\[\sin \left(\frac{\pi}{2}-x\right)=\cos (x)\]
and
\[\cos \left(\frac{\pi}{2}-x\right)=\sin (x).\]
Show that for any \(x \in \mathbb{R}\),
\[\sin (2 x)=2 \sin (x) \cos (x)\]
and
\[\cos (2 x)=\cos ^{2}(x)-\sin ^{2}(x).\]
Show that for any \(x \in \mathbb{R}\),
\[\sin ^{2}(x)=\frac{1-\cos (2 x)}{2}\]
and
\[\cos ^{2}(x)=\frac{1+\cos (2 x)}{2}.\]
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
\(\lim _{x \rightarrow 0} \frac{\arctan (x)}{x}=1\).
- 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
\(\lim _{x \rightarrow 0} \frac{\tan (x)}{x}=1\).
- 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
\(\lim _{x \rightarrow 0} \frac{\sin (x)}{x}=1\).
- Proof
-
We have
\[\lim _{x \rightarrow 0} \frac{\sin (x)}{x}=\lim _{x \rightarrow 0} \frac{\tan (x)}{x} \cos (x)=1.\]
\(\lim _{x \rightarrow 0} \frac{1-\cos (x)}{x}=0 .\)
- 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
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
If \(f(x)=\cos (x),\) then \(f^{\prime}(x)=-\sin (x)\).
Prove the previous proposition.
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.
If \(f(x)=\tan (x)\) and \(g(x)=\cot (x),\) show that
\[f^{\prime}(x)=\sec ^{2}(x)\]
and
\[g^{\prime}(x)=-\csc ^{2}(x).\]
If \(f(x)=\sec (x)\) and \(g(x)=\csc (x),\) show that
\[f^{\prime}(x)=\sec (x) \tan (x),\]
and
\[g^{\prime}(x)=-\csc (x) \cot (x).\]
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.
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.