# Sine function

(Redirected from Sin)
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 $\R$
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 $\sin^2x + \cos^2x = 1$.
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 $\cos(2x) = 1 - 2\sin^2x$, so $\sin^2x = (1 - \cos(2x))/2$.
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 $|\sin|$ 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 $x \mapsto \ln|\sin x|$ 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 $\sin' = \cos$ 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

$\cos(x_0)$

### 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:

$\int \sin x \, dx = -\cos x + C$

### Definite integrals

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

$\int_a^{a + 2\pi} \sin x \, dx = 0$

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

$\int_{-a}^a \sin x \, dx = 0$

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:

$\int_0^{\pi/2} \sin x \, dx = \int_{\pi/2}^\pi \sin x \, dx = 1$

### Integration of transformed versions of function

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

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

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

### Higher antiderivatives

The general expression for the second antiderivative is:

$\int \int \sin x \, dx \, dx = -\sin x + C_1x + C_0$

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 $k - 1$.

### Integration of product with polynomials

Using integration by parts, we know that knowledge of the first $k$ antiderivatives of $f$ is sufficient to determine $\int x^{k-1} f(x) \, dx$ via repeated application of integration by parts. Since the sine function can be antidifferentiated any number of times, this allows us to antidifferentiate any polynomial times the sine function.

For instance, the function $x \mapsto x \sin x$, i.e., the product of identity function and sine function, can be integrated via knowledge of how to integrate the sine function twice:

$\int x \sin x \, dx = x \int \sin x \, dx - \int \int \sin x \, dx \, dx = -x\cos x + \sin x + C$

## Differential equations

The sine function and its transforms arises as the solution to many differential equations, including polynomial differential equations. Some of these are listed below.

Equation Other forms General solution
$\frac{dx}{dt} = \sqrt{a^2 - x^2}, \qquad a > 0$ $x = a \sin (t + \varphi)$ for $-\pi/2 \le t + \varphi \le \pi/2$
$\frac{dx}{dt} = a \sqrt{x(b- x)}, \qquad a,b > 0$ $\left(\frac{dx}{dt}\right)^2 = a^2x(b- x), \qquad a,b > 0$ See Verhulst process
$\frac{d^2x}{dt^2} = -x$ $x = a \sin(t + \varphi)$ or $x = A \sin t + B \cos t$

## 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:

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

### Taylor series equals power series

The sine function is a globally analytic function: the Taylor series for the sine function does in fact converge to the function everywhere. This can be proved in a number of ways. One method is to use that uniformly bounded derivatives implies analytic. Alternatively, we could note that $\sin$ satisfies a certain differential equation $y'' = -y$, forcing it to be given by a power series.

Thus we have that:

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

## Limit computations

### Order of zero

We have the following limit:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

Thus, the order of zero at 0 is 1 and the residue is 1.

### Higher order limits

We have the limit:

$\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac{1}{6}$

The limit can be computed in either of two ways:

Name of method for computing the limit Details
Using the L'Hopital rule $\lim_{x \to 0} \frac{x - \sin x}{x^3} \stackrel{*}{=} \lim_{x \to 0} \frac{1 -\cos x}{3x^2} \stackrel{*}{=} \lim_{x \to 0} \frac{\sin x}{6x} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x}{6} = 1/6$.
Using the power series We have $\sin x = x - x^3/3! + O(x^5)$, so $\frac{x - \sin x}{x^3} = \frac{1}{3!} - O(x^2)$, so the limit as $x \to 0$ is $1/3! = 1/6$.