Sine function

This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.
View a complete list of particular functions on this wiki

For functions involving angles (trigonometric functions, inverse trigonometric functions, etc.) we follow the convention that all angles are measured in radians. Thus, for instance, the angle of

$90{}^{\circ }$

is measured as

$\pi /2$

.

Definition

Unit circle definition

The sine function, denoted $\sin$, is defined as follows.

Consider the unit circle centered at the origin, described as the following subset of the coordinate:

$\{(x,y)\mid x^2+y^2=1\}$

For a real number $t$, we define $\sin t$ as follows:

• Start at the point $(1,0)$, which lies on the unit circle centered at the origin.
• Move a distance of $t$ along the unit circle in the counter-clockwise direction (i.e., the motion begins in the first quadrant, with both coordinates positive).
• At the end, the $y$-coordinate of the point thus obtained is defined as $\sin t$.

Triangle ratio definition (works for acute angles)

For an acute angle $t$, i.e., for $t$ in the open interval $(0,\pi /2)$, $\sin t$ can be defined as follows:

• Construct any right triangle with one of the acute angles equal to $t$.
• $\sin t$ is the ratio of the leg opposite to the angle $t$ to the hypotenuse.

Key data

Item	Value
default domain	all real numbers, i.e., all of $\mathbb{R}$
range	the closed interval $[-1,1]$
period	$2\pi$, i.e., $360{}^{\circ }$.
mean value over a period	0
local maximum values and points of attainment	local maximum value attained at all points of the form $2n\pi +\pi /2,n\in \mathbb{Z}$, with value 1 at each point.
local minimum values and points of attainment	local minimum value attained at all points of the form $2n\pi -\pi /2,n\in \mathbb{Z}$, with value -1 at each point.
points of inflection (both coordinates)	all points of the form $(n\pi ,0)$ with $n\in \mathbb{Z}$.
important symmetries	odd function. More generally, half turn symmetry about all points of the form $(n\pi ,0)$ where $n\in \mathbb{Z}$. Also, mirror symmetry about all lines of the form $x=n\pi +\pi /2$.
first derivative	$\cos$, i.e., the cosine function
second derivative	$-\sin$, i.e., the negative of the sine function.
sequence of derivatives	starting from first: $\cos ,-\sin ,-\cos ,\sin ,\cos ,-\sin ,-\cos ,\sin ,\dots$. The sequence of higher derivatives is periodic with a period of 4.
first antiderivative	$-\cos$, i.e., the negative of the cosine function.

Identities

Type of identity	Identity in algebraic form
complementary angle	$\sin (\pi /2-x)=\cos x$, equivalently, $\cos (\pi /2-x)=\sin x$
square relationship with cosine	${\sin }^{2}x+{\cos }^{2}x=1$.
angle sum sine formula	$\sin (x+y)=\sin x\cos y+\cos x\sin y$
angle difference sine formula	$\sin (x-y)=\sin x\cos y-\cos x\sin y$
product to sum conversion	$\sin x\sin y=\frac{1}{2}(\cos (x-y)-\cos (x+y))$
sum to product conversion	$\sin x+\sin y=2\sin (\frac{x+y}{2})\cos (\frac{x-y}{2})$
double angle sine formula	$\sin (2x)=2\sin x\cos x$
double angle cosine formula	$\cos (2x)=1-2{\sin }^{2}x$, so ${\sin }^{2}x=(1-\cos (2x))/2$.
other symmetries	periodicity: $\sin (2\pi +x)=\sin x$ anti-periodicity: $\sin (\pi +x)=-\sin x$ odd: $\sin (-x)=-\sin x$ mirror symmetry about $\pi /2$: $\sin (\pi -x)=\sin x$

Related functions

Composition with other functions

Below are some composite functions of the form $f\circ \sin$ for suitable function $f$:

Function $f$	$f\circ \sin$, i.e., the function $x\mapsto f(\sin x)$	$\sin \circ f$ i.e. the function $x\mapsto \sin (f(x))$
reciprocal function	cosecant function $\csc$ -- the domain of this excludes all multiples of $\pi$	sine of reciprocal function $x\mapsto \sin (1/x)$
square function	sine-squared function ${\sin }^2$	sine of square function $x\mapsto \sin (x^2)$
cube function	sine-cubed function ${\sin }^3$	sine of cube function $x\mapsto \sin (x^3)$
positive power function $x\mapsto x^n$ for fixed $n$	positive power of sine function	sine of positive power function
absolute value function	absolute value of sine function $\left|\sin \right|$	uninteresting
positive part function	positive part of sine function ${\sin }^{+}$	uninteresting
square root function	square root of sine function -- note that this makes sense only on a restricted domain	sine of square root function
natural logarithm of absolute value	natural logarithm of absolute value of sine function $x\mapsto \ln \left|\sin x\right|$	sine of natural logarithm of absolute value $x\mapsto \sin (\ln |x|)$

Product with other functions

Function $f$	pointwise product of functions $x\mapsto f(x)(\sin x)$
reciprocal function	sinc function $x\mapsto (\sin x)/x$ (separately defined as 1 at 0)
identity function	product of identity function and sine function $x\mapsto x\sin x$
exponential function	product of exponential function and sine function $x\mapsto e^x\sin x$
cosine function	$x\mapsto \sin x\cos x$ is same as $x\mapsto (1/2)\sin (2x)$

Differentiation

First derivative

We deduce the formula ${{\sin }^{\prime }}^{\prime }=\cos$ from the limit:

$\underset{h\to 0}{lim}\frac{\sin h}{h}=1$

Here's the full proof:

${{\sin }^{\prime }}^{\prime }(x_0)=\underset{x\to x_0}{lim}\frac{\sin x-\sin (x_0)}{x-x_0}\stackrel{h:=x-x_0}{=}\underset{h\to 0}{lim}\frac{\sin (x_0+h)-\sin x_0}{h}=\underset{h\to 0}{lim}\frac{\sin (x_0)\cos h+\cos (x_0)\sin h-\sin (x_0)}{h}$

By the fact that limit is linear, the above limit can be rewritten as:

$\sin (x_0)(\underset{h\to 0}{lim}\frac{\cos h-1}{h})+\cos (x_0)(\underset{h\to 0}{lim}\frac{\sin h}{h})$

We now need to compute the two limits individually. Note first that both limits are independent of $x_0$.

The first limit is:

$\underset{h\to 0}{lim}\frac{\cos h-1}{h}=\underset{h\to 0}{lim}\frac{-2{\sin }^2(h/2)}{h}=\underset{h\to 0}{lim}-\sin (h/2)\underset{h\to 0}{lim}\frac{\sin (h/2)}{(h/2)}$

We've thus expressed the limit as a product of limits where one of the factors goes to zero and the other goes to one, so the limit is zero.

The second limit is 1, as can be seen directly.

We thus get that the answer is:

$\sin (x_0)(0)+\cos (x_0)(1)$

This simplifies to

$\cos (x_0)$

Second derivative

The derivative of $\cos$ is $-\sin$, so we obtain:

${{\sin }^{″}}^{″}=-\sin$

Higher derivatives

The sequence of derivatives is periodic with period 4:

$k$	0	1	2	3	4	5	6	7
${\sin }^{(k)}$	$\sin$	$\cos$	$-\sin$	$-\cos$	$\sin$	$\cos$	$-\sin$	$-\cos$

In particular, we obtain that for any nonnegative integer $k$:

• ${\sin }^{(4k)}=\sin$
• ${\sin }^{(4k+1)}=\cos$
• ${\sin }^{(4k+2)}=-\sin$
• ${\sin }^{(4k+3)}=-\cos$

Equivalently, we also have:

${\sin }^{(k)}(x)=\sin (x+k\pi /2)$

In other words, differentiating the function $k$ times is equivalent to shifting the graph $k\pi /2$ to the left.

Integration

First antiderivative

We have:

$\int \sin xdx=-\cos x+C$

Definite integrals

The mean value of $\sin$ over a period is 0. Thus:

$\underset{a}{\overset{a+2\pi }{\int }}\sin xdx=0$

Since $\sin$ is odd, the mean value over any interval symmetric about the origin is zero:

$\underset{-a}{\overset{a}{\int }}\sin xdx=0$

Also, the integral of $\sin$ on $[0,\pi /2]$ and on $[\pi /2,\pi ]$ is 1 each, giving a mean value of $2/\pi$ on these intervals:

$\underset{0}{\overset{\pi /2}{\int }}\sin xdx=\underset{\pi /2}{\overset{\pi }{\int }}\sin xdx=1$

Integration of transformed versions of function

We have, for $m\ne 0$, the following, using integration of linear transform of function:

$\int \sin (mx+\phi )dx=\frac{-1}{m}\cos (mx+\phi )+C$

Further, the mean value of $\sin (mx+\phi )$ over a period is 0.

Higher antiderivatives

The general expression for the second antiderivative is:

$\int \int \sin xdxdx=-\sin x+C_{1}x+C_0$

In general, the $k^{th}$ antiderivative is $\pm \cos$ or $\pm \sin$, depending on the value of $k$ mod 4. The general expression is the particular antiderivative plus an arbitrary polynomial of degree at most $k-1$.

Power series and Taylor series

Computation of Taylor series

As noted above, we have that:

${\sin }^{(k)}x=\sin (x+k\pi /2)$

In particular, this means that:

${\sin }^{(k)}0=\sin (k\pi /2)$

Thus, the sequence of derivatives at zero (starting from $k=0$) is $0,1,0,-1,0,1,0,-1,\dots$.

The Taylor series is thus:

$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots =\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{2k+1}}{(2k+1)!}$

Taylor series equals power series

There are a number of ways of showing that the Taylor series for the sine function does in fact converge to the function everywhere. One of these follows from the max-estimate on the remainder term from the Taylor polyonmials. Since the sine function is a bounded function, this max-estimate approaches zero as $n\to \mathrm{\infty }$.

Thus we have that:

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots =\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{2k+1}}{(2k+1)!}$