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	$\sin x\sin y={\frac {1}{2}}\left(\cos(x-y)-\cos(x+y)\right)$
sum to product conversion	$\sin x+\sin y=2\sin \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)$
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 $\pi /2$: $\sin(\pi -x)=\sin x$

Differentiation

We deduce the formula $\sin '=\cos$ from the limit:

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

Here's the full proof:

$\!\sin '(x_{0})=\lim _{x\to x_{0}}{\frac {\sin x-\sin(x_{0})}{x-x_{0}}}{\stackrel {h:=x-x_{0}}{=}}\lim _{h\to 0}{\frac {\sin(x_{0}+h)-\sin x_{0}}{h}}=\lim _{h\to 0}{\frac {\sin(x_{0})\cos h+\cos(x_{0})\sin h-\sin(x_{0})}{h}}$

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

$\!\sin(x_{0})\left(\lim _{h\to 0}{\frac {\cos h-1}{h}}\right)+\cos(x_{0})\left(\lim _{h\to 0}{\frac {\sin h}{h}}\right)$

We now need to compute the two limits individually. Note first that both limits are independent of $x_{0}$.

The first limit is:

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

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

$\cos(x_{0})$