# Proofs of trigonometric identities

Proofs of trigonometric identities are used to show relations between trigonometric functions. This article will list trigonometric identities and prove them.

## Elementary trigonometric identities

### Definitions

Trigonometric functions specify the relationships between side lengths and interior angles of a right triangle. For example, the sine of angle θ is defined as being the length of the opposite side divided by the length of the hypotenuse.

Referring to the diagram at the right, the six trigonometric functions of θ are:

$\sin \theta = \frac {\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac {a}{h}$
$\cos \theta = \frac {\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac {b}{h}$
$\tan \theta = \frac {\mathrm{opposite}}{\mathrm{adjacent}} = \frac {a}{b}$
$\cot \theta = \frac {\mathrm{adjacent}}{\mathrm{opposite}} = \frac {b}{a}$
$\sec \theta = \frac {\mathrm{hypotenuse}}{\mathrm{adjacent}} = \frac {h}{b}$
$\csc \theta = \frac {\mathrm{hypotenuse}}{\mathrm{opposite}} = \frac {h}{a}$

### Ratio identities

The following identities are trivial algebraic consequences of these definitions and the division identity.
They rely on multiplying or dividing the numerator and denominator of fractions by a variable. Ie,

$\frac {a}{b}= \frac {\left(\frac {a}{c}\right)} {\left(\frac {b}{c}\right) }$
$\tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} = \frac { \left( \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} \right) } { \left( \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}\right) } = \frac {\sin \theta} {\cos \theta}$
$\cot \theta =\frac{\mathrm{adjacent}}{\mathrm{opposite}} = \frac { \left( \frac{\mathrm{adjacent}}{\mathrm{adjacent}} \right) } { \left( \frac {\mathrm{opposite}}{\mathrm{adjacent}} \right) } = \frac {1}{\tan \theta} = \frac {\cos \theta}{\sin \theta}$
$\sec \theta = \frac {1}{\cos \theta} = \frac{\mathrm{hypotenuse}}{\mathrm{adjacent}}$
$\csc \theta = \frac {1}{\sin \theta} = \frac{\mathrm{hypotenuse}}{\mathrm{opposite}}$
$\tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} = \frac{\left(\frac{\mathrm{opposite} \times \mathrm{hypotenuse}}{\mathrm{opposite} \times \mathrm{adjacent}} \right) } { \left( \frac {\mathrm{adjacent} \times \mathrm{hypotenuse}} {\mathrm{opposite} \times \mathrm{adjacent} } \right) } = \frac{\left( \frac{\mathrm{hypotenuse}}{\mathrm{adjacent}} \right)} { \left( \frac{\mathrm{hypotenuse}}{\mathrm{opposite}} \right)} = \frac {\sec \theta}{\csc \theta}$

Or

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\left( \frac{1}{\csc \theta} \right) }{\left( \frac{1}{\sec \theta} \right) } = \frac{\left( \frac{\csc \theta \sec \theta}{\csc \theta} \right) }{\left( \frac{\csc \theta \sec \theta}{\sec \theta} \right) } = \frac{\sec \theta}{\csc \theta}$
$\cot \theta = \frac {\csc \theta}{\sec \theta}$

### Complementary angle identities

Two angles whose sum is π/2 radians (90 degrees) are complementary. In the diagram, the angles at vertices A and B are complementary, so we can exchange a and b, and change θ to π/2 − θ, obtaining:

$\sin\left( \pi/2-\theta\right) = \cos \theta$
$\cos\left( \pi/2-\theta\right) = \sin \theta$
$\tan\left( \pi/2-\theta\right) = \cot \theta$
$\cot\left( \pi/2-\theta\right) = \tan \theta$
$\sec\left( \pi/2-\theta\right) = \csc \theta$
$\csc\left( \pi/2-\theta\right) = \sec \theta$

### Pythagorean identities

Identity 1:

$\sin^2(x) + \cos^2(x) = 1\,$

Proof 1:

Refer to the triangle diagram above. Note that $a^2+b^2=h^2$ by Pythagorean theorem.

$\sin^2(x) + \cos^2(x) = \frac{a^2}{h^2} + \frac{b^2}{h^2} = \frac{a^2+b^2}{h^2} = \frac{h^2}{h^2} = 1.\,$

The following two results follow from this and the ratio identities. To obtain the first, divide both sides of $\sin^2(x) + \cos^2(x) = 1$ by $\cos^2(x)$; for the second, divide by $\sin^2(x)$.

$\tan^2(x) + 1\ = \sec^2(x)$
$\sec^2(x) - \tan^2(x) = 1\$

Similarly

$1\ + \cot^2(x) = \csc^2(x)$
$\csc^2(x) - \cot^2(x) = 1\$

Proof 2:

Differentiating the left-hand side of the identity yields:

$2 \sin x \cdot \cos x - 2 \sin x \cdot \cos x = 0$

Integrating this shows that the original identity is equal to a constant, and this constant can be found by plugging in any arbitrary value of x.

Identity 2:

The following accounts for all three reciprocal functions.

$\csc^2(x) + \sec^2(x) - \cot^2(x) = 2\ + \tan^2(x)$

Proof 1:

Refer to the triangle diagram above. Note that $a^2+b^2=h^2$ by Pythagorean theorem.

$\csc^2(x) + \sec^2(x) = \frac{h^2}{a^2} + \frac{h^2}{b^2} = \frac{a^2+b^2}{a^2} + \frac{a^2+b^2}{b^2} = 2\ + \frac{b^2}{a^2} + \frac{a^2}{b^2}$

Substituting with appropriate functions -

$2\ + \frac{b^2}{a^2} + \frac{a^2}{b^2} = 2\ + \tan^2(x)+ \cot^2(x)$

Rearranging gives:

$\csc^2(x) + \sec^2(x) - \cot^2(x) = 2\ + \tan^2(x)$

### Angle sum identities

#### Sine

Illustration of the sum formula.

Draw a horizontal line (the x-axis); mark an origin O. Draw a line from O at an angle $\alpha$ above the horizontal line and a second line at an angle $\beta$ above that; the angle between the second line and the x-axis is $\alpha + \beta$.

Place P on the line defined by $\alpha + \beta$ at a unit distance from the origin.

Let PQ be a line perpendicular to line defined by angle $\alpha$, drawn from point Q on this line to point P. $\therefore$ OQP is a right angle.

Let QA be a perpendicular from point A on the x-axis to Q and PB be a perpendicular from point B on the x-axis to P. $\therefore$ OAQ and OBP are right angles.

Draw QR parallel to the x-axis.

Now angle $RPQ = \alpha$ (because $OQA = 90 - \alpha$, making $RQO = \alpha, RQP = 90-\alpha$, and finally $RPQ = \alpha$)

$RPQ = \tfrac{\pi}{2} - RQP = \tfrac{\pi}{2} - (\tfrac{\pi}{2} - RQO) = RQO = \alpha$
$OP = 1$
$PQ = \sin \beta$
$OQ = \cos \beta$
$\frac{AQ}{OQ} = \sin \alpha\,$, so $AQ = \sin \alpha \cos \beta$
$\frac{PR}{PQ} = \cos \alpha\,$, so $PR = \cos \alpha \sin \beta$
$\sin (\alpha + \beta) = PB = RB+PR = AQ+PR = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

By substituting $-\beta$ for $\beta$ and using Symmetry, we also get:

$\sin (\alpha - \beta) = \sin \alpha \cos -\beta + \cos \alpha \sin -\beta$
$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

Another simple "proof" can be given using Euler's formula known from complex analysis: Euler's formula is:

$e^{i\varphi}=\cos \varphi +i \sin \varphi$

Although it is more precise to say that Euler's formula entails the trigonometric identities, it follows that for angles $\alpha$ and $\beta$ we have:

$e^{i (\alpha + \beta)} = \cos (\alpha +\beta) + i \sin(\alpha +\beta)$

Also using the following properties of exponential functions:

$e^{i(\alpha + \beta)} = e^{i \alpha} e^{i\beta}= (\cos \alpha +i \sin \alpha) (\cos \beta + i \sin \beta)$

Evaluating the product:

$e^{i(\alpha + \beta)} = (\cos \alpha \cos \beta - \sin \alpha \sin \beta)+i(\sin \alpha \cos \beta + \sin \beta \cos \alpha)$

Equating real and imaginary parts:

$\cos (\alpha +\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta$
$\sin (\alpha +\beta)=\sin \alpha \cos \beta + \sin \beta \cos \alpha$

#### Cosine

Using the figure above,

$OP = 1\,$
$PQ = \sin \beta\,$
$OQ = \cos \beta\,$
$\frac{OA}{OQ} = \cos \alpha\,$, so $OA = \cos \alpha \cos \beta\,$
$\frac{RQ}{PQ} = \sin \alpha\,$, so $RQ = \sin \alpha \sin \beta\,$
$\cos (\alpha + \beta) = OB = OA-BA = OA-RQ = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta\,$

By substituting $-\beta$ for $\beta$ and using Symmetry, we also get:

$\cos (\alpha - \beta) = \cos \alpha \cos - \beta\ - \sin \alpha \sin - \beta\,$
$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\,$

Also, using the complementary angle formulae,

\begin{align} \cos (\alpha + \beta) & = \sin\left( \pi/2-(\alpha + \beta)\right) \\ & = \sin\left( (\pi/2-\alpha) - \beta\right) \\ & = \sin\left( \pi/2-\alpha\right) \cos \beta - \cos\left( \pi/2-\alpha\right) \sin \beta \\ & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \end{align}

#### Tangent and cotangent

From the sine and cosine formulae, we get

$\tan (\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}$

Dividing both numerator and denominator by $\cos \alpha \cos \beta$, we get

$\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

Subtracting $\beta$ from $\alpha$, using $\tan (- \beta) = -\tan \beta$,

$\tan (\alpha - \beta) = \frac{\tan \alpha + \tan (-\beta)}{1 - \tan \alpha \tan (-\beta)} = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

Similarly from the sine and cosine formulae, we get

$\cot (\alpha + \beta) = \frac{\cos (\alpha + \beta)}{\sin (\alpha + \beta)} = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \alpha \cos \beta + \cos \alpha \sin \beta}$

Then by dividing both numerator and denominator by $\sin \alpha \sin \beta$, we get

$\cot (\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$

Or, using $\cot \theta = \frac{1}{\tan \theta}$,

$\cot (\alpha + \beta) = \frac{1 - \tan \alpha \tan \beta}{\tan \alpha + \tan \beta} = \frac{\frac{1}{\tan \alpha \tan \beta} - 1}{\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}} = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$

Using $\cot (- \beta) = -\cot \beta$,

$\cot (\alpha - \beta) = \frac{\cot \alpha \cot (-\beta) - 1}{ \cot \alpha + \cot (-\beta) } = \frac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}$

### Double-angle identities

From the angle sum identities, we get

$\sin (2 \theta) = 2 \sin \theta \cos \theta\,$

and

$\cos (2 \theta) = \cos^2 \theta - \sin^2 \theta\,$

The Pythagorean identities give the two alternative forms for the latter of these:

$\cos (2 \theta) = 2 \cos^2 \theta - 1\,$
$\cos (2 \theta) = 1 - 2 \sin^2 \theta\,$

The angle sum identities also give

$\tan (2 \theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2}{\cot \theta - \tan \theta}\,$
$\cot (2 \theta) = \frac{\cot^2 \theta - 1}{2 \cot \theta} = \frac{\cot \theta - \tan \theta}{2}\,$

It can also be proved using Euler's formula

$e^{i \varphi}=\cos \varphi +i \sin \varphi$

Squaring both sides yields

$e^{i 2\varphi}=(\cos \varphi +i \sin \varphi)^{2}$

But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields

$e^{i 2\varphi}=\cos 2\varphi +i \sin 2\varphi$

It follows that

$(\cos \varphi +i \sin \varphi)^{2}=\cos 2\varphi +i \sin 2\varphi$.

Expanding the square and simplifying on the left hand side of the equation gives

$i(2 \sin \varphi \cos \varphi) + \cos^2 \varphi - \sin^2 \varphi\ = \cos 2\varphi +i \sin 2\varphi$.

Because the imaginary and real parts have to be the same, we are left with the original identities

$\cos^2 \varphi - \sin^2 \varphi\ = \cos 2\varphi$,

and also

$2 \sin \varphi \cos \varphi = \sin 2\varphi$.

### Half-angle identities

The two identities giving the alternative forms for cos 2θ lead to the following equations:

$\cos \frac{\theta}{2} = \pm\, \sqrt\frac{1 + \cos \theta}{2},\,$
$\sin \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{2}.\,$

The sign of the square root needs to be chosen properly—note that if π is added to θ, the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore the correct sign to use depends on the value of θ.

For the tan function, the equation is:

$\tan \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{1 + \cos \theta}.\,$

Then multiplying the numerator and denominator inside the square root by (1 + cos θ) and using Pythagorean identities leads to:

$\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}.\,$

Also, if the numerator and denominator are both multiplied by (1 - cos θ), the result is:

$\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}.\,$

This also gives:

$\tan \frac{\theta}{2} = \csc \theta - \cot \theta.\,$

Similar manipulations for the cot function give:

$\cot \frac{\theta}{2} = \pm\, \sqrt\frac{1 + \cos \theta}{1 - \cos \theta} = \frac{1 + \cos \theta}{\sin \theta} = \frac{\sin \theta}{1 - \cos \theta} = \csc \theta + \cot \theta.\,$

### Miscellaneous -- the triple tangent identity

If $\psi$, $\theta$ and $\phi$ are the angles of a triangle, i.e. $\psi + \theta + \phi = \pi =$ half circle,

$\tan(\psi) + \tan(\theta) + \tan(\phi) = \tan(\psi)\tan(\theta)\tan(\phi).$

Proof:[1]

\begin{align} \psi & = \pi - \theta - \phi \\ \tan(\psi) & = \tan(\pi - \theta - \phi) \\ & = - \tan(\theta + \phi) \\ & = \frac{- \tan\theta - \tan\phi}{1 - \tan\theta \tan\phi} \\ & = \frac{\tan\theta + \tan\phi}{\tan\theta \tan\phi - 1} \\ (\tan\theta \tan\phi - 1) \tan\psi & = \tan\theta + \tan\phi \\ \tan\psi \tan\theta \tan\phi - \tan\psi & = \tan\theta + \tan\phi \\ \tan\psi \tan\theta \tan\phi & = \tan\psi + \tan\theta + \tan\phi \\ \end{align}

### Miscellaneous -- the triple cotangent identity

If $\psi + \theta + \phi = \tfrac{\pi}{2} =$ quarter circle,

$\cot(\psi) + \cot(\theta) + \cot(\phi) = \cot(\psi)\cot(\theta)\cot(\phi)$.

Proof:

Replace each of $\psi$, $\theta$, and $\phi$ with their complementary angles, so cotangents turn into tangents and vice-versa.

Given

$\psi + \theta + \phi = \tfrac{\pi}{2}\,$
$\therefore (\tfrac{\pi}{2}-\psi) + (\tfrac{\pi}{2}-\theta) + (\tfrac{\pi}{2}-\phi) = \tfrac{3\pi}{2} - (\psi+\theta+\phi) = \tfrac{3\pi}{2} - \tfrac{\pi}{2} = \pi$

so the result follows from the triple tangent identity.

### Prosthaphaeresis identities

• $\sin \theta \pm \sin \phi = 2 \sin \left ( \frac{\theta\pm \phi}2 \right ) \cos \left ( \frac{\theta\mp \phi}2 \right )$
• $\cos \theta + \cos \phi = 2 \cos \left ( \frac{\theta+\phi}2 \right ) \cos \left ( \frac{\theta-\phi}2 \right )$
• $\cos \theta - \cos \phi = -2 \sin \left ( \frac{\theta+\phi}2 \right ) \sin \left ( \frac{\theta-\phi}2 \right )$

#### Proof of sine identities

$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

By adding these together,

$\sin (\alpha + \beta) + \sin (\alpha - \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta + \sin \alpha \cos \beta - \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta$

Similarly, by subtracting the two sum-angle identities,

$\sin (\alpha + \beta) - \sin (\alpha - \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta - \sin \alpha \cos \beta + \cos \alpha \sin \beta = 2 \cos \alpha \sin \beta$

Let $\alpha + \beta = \theta$ and $\alpha - \beta = \phi$,

$\therefore \alpha = \frac{\theta + \phi}2$ and $\beta = \frac{\theta - \phi}2$

Substitute $\theta$ and $\phi$

$\sin \theta + \sin \phi = 2 \sin \left( \frac{\theta + \phi}2 \right) \cos \left( \frac{\theta - \phi}2 \right)$
$\sin \theta - \sin \phi = 2 \cos \left( \frac{\theta + \phi}2 \right) \sin \left( \frac{\theta - \phi}2 \right) = 2 \sin \left( \frac{\theta - \phi}2 \right) \cos \left( \frac{\theta + \phi}2 \right)$

Therefore,

$\sin \theta \pm \sin \phi = 2 \sin \left( \frac{\theta\pm \phi}2 \right) \cos \left( \frac{\theta\mp \phi}2 \right)$

#### Proof of cosine identities

Similarly for cosine, start with the sum-angle identities:

$\cos (\alpha + \beta) = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta$
$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Again, by adding and subtracting

$\cos (\alpha + \beta) + \cos (\alpha - \beta) = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta + \cos \alpha \cos \beta + \sin \alpha \sin \beta = 2\cos \alpha \cos \beta\$
$\cos (\alpha + \beta) - \cos (\alpha - \beta) = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta - \cos \alpha \cos \beta - \sin \alpha \sin \beta = -2 \sin \alpha \sin \beta$

Substitute $\theta$ and $\phi$ as before,

$\cos \theta + \cos \phi = 2 \cos \left( \frac{\theta+\phi}2 \right) \cos \left( \frac{\theta-\phi}2 \right)$
$\cos \theta - \cos \phi = -2 \sin \left( \frac{\theta+\phi}2 \right) \sin \left( \frac{\theta-\phi}2 \right)$

### Inequalities

Illustration of the sine and tangent inequalities.

The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2π) of the whole circle, so its area is θ/2.

$OA = OD = 1\,$
$AB = \sin \theta\,$
$CD = \tan \theta\,$

The area of triangle OAD is AB/2, or sinθ/2. The area of triangle OCD is CD/2, or tanθ/2.

Since triangle OAD lies completely inside the sector, which in turn lies completely inside triangle OCD, we have

$\sin \theta < \theta < \tan \theta\,$

This geometric argument applies if 0<θ<π/2. It relies on definitions of arc length and area, which act as assumptions, so it is rather a condition imposed in construction of trigonometric functions than a provable property.[2] For the sine function, we can handle other values. If θ>π/2, then θ>1. But sinθ≤1 (because of the Pythagorean identity), so sinθ<θ. So we have

$\frac{\sin \theta}{\theta} < 1\ \ \ \mathrm{if}\ \ \ 0 < \theta\,$

For negative values of θ we have, by symmetry of the sine function

$\frac{\sin \theta}{\theta} = \frac{\sin (-\theta)}{-\theta} < 1\,$

Hence

$\frac{\sin \theta}{\theta} < 1\ \ \ \mathrm{if}\ \ \ \theta \ne 0\,$
$\frac{\tan \theta}{\theta} > 1\ \ \ \mathrm{if}\ \ \ 0 < \theta < \frac{\pi}{2}\,$

## Identities involving calculus

### Preliminaries

$\lim_{\theta \to 0}{\sin \theta} = 0\,$
$\lim_{\theta \to 0}{\cos \theta} = 1\,$

### Sine and angle ratio identity

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

Proof: From the previous inequalities, we have, for small angles

$\sin \theta < \theta < \tan \theta\,$,

Therefore,

$\frac{\sin \theta}{\theta} < 1 < \frac{\tan \theta}{\theta}\,$,

Consider the right-hand inequality. Since

$\tan \theta = \frac{\sin \theta}{\cos \theta}$
$\therefore 1 < \frac{\sin \theta}{\theta \cos \theta}$

Multply through by $\cos \theta$

$\cos \theta < \frac{\sin \theta}{\theta}$

Combining with the left-hand inequality:

$\cos \theta < \frac{\sin \theta}{\theta} < 1$

Taking $\cos \theta$ to the limit as $\theta \to 0$

$\lim_{\theta \to 0}{\cos \theta} = 1\,$

Therefore,

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

### Cosine and angle ratio identity

$\lim_{\theta \to 0}\frac{1 - \cos \theta}{\theta} = 0$

Proof:

\begin{align} \frac{1 - \cos \theta}{\theta} & = \frac{1 - \cos^2 \theta}{\theta (1 + \cos \theta)}\\ & = \frac{\sin^2 \theta}{\theta (1 + \cos \theta)}\\ & = \left( \frac{\sin \theta}{\theta} \right) \times \sin \theta \times \left( \frac{1}{1 + \cos \theta} \right)\\ \end{align}

The limits of those three quantities are 1, 0, and 1/2, so the resultant limit is zero.

### Cosine and square of angle ratio identity

$\lim_{\theta \to 0}\frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$

Proof:

As in the preceding proof,

$\frac{1 - \cos \theta}{\theta^2} = \frac{\sin \theta}{\theta} \times \frac{\sin \theta}{\theta} \times \frac{1}{1 + \cos \theta}.\,$

The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2.

### Proof of Compositions of trig and inverse trig functions

All these functions follow from the Pythagorean trigonometric identity. We can prove for instance the function

$\sin[\arctan(x)]=\frac{x}{\sqrt{1+x^2}}$

Proof:

We start from

$\sin^2\theta+\cos^2\theta=1$

Then we divide this equation by $\cos^2\theta$

$\cos^2\theta=\frac{1}{\tan^2\theta+1}$

Then use the substitution $\theta=\arctan(x)$, also use the Pythagorean trigonometric identity:

$1-\sin^2[\arctan(x)]=\frac{1}{\tan^2[\arctan(x)]+1}$

Then we use the identity $\tan[\arctan(x)]\equiv x$

$\sin[\arctan(x)]=\frac{x}{\sqrt{x^2+1}}$