View source for Proofs of trigonometric identities

{{Short description|Collection of proofs of equations involving trigonometric functions}}

There are several equivalent ways for defining [[trigonometric function]]s, and the proofs of the [[trigonometric identities]] between them depend on the chosen definition. The oldest and most elementary definitions are based on the geometry of [[right triangle]]s and the ratio between their sides. The proofs given in this article use these definitions, and thus apply to non-negative angles not greater than a [[right angle]]. For greater and negative [[angle]]s, see [[Trigonometric functions]].

Other definitions, and therefore other proofs are based on the [[Taylor series]] of [[sine]] and [[cosine]], or on the [[differential equation]] <math>f''+f=0</math> to which they are solutions.

==Elementary trigonometric identities==

===Definitions===

[[File: Trigonometry triangle.svg|thumb|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.]]

The six trigonometric functions are defined for every [[real number]], except, for some of them, for angles that differ from 0 by a multiple of the right angle (90°). Referring to the diagram at the right, the six trigonometric functions of θ are, for angles smaller than the right angle:

:<math> \sin \theta = \frac {\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac {a}{h}</math>

:<math> \cos \theta = \frac {\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac {b}{h}</math>

:<math> \tan \theta = \frac {\mathrm{opposite}}{\mathrm{adjacent}} = \frac {a}{b}</math>

:<math> \cot \theta = \frac {\mathrm{adjacent}}{\mathrm{opposite}} = \frac {b}{a}</math>

:<math> \sec \theta = \frac {\mathrm{hypotenuse}}{\mathrm{adjacent}} = \frac {h}{b}</math>

:<math> \csc \theta = \frac {\mathrm{hypotenuse}}{\mathrm{opposite}} = \frac {h}{a}</math>

===Ratio identities===

In the case of angles smaller than a right angle, the following identities are direct consequences of above definitions through the division identity
: <math> \frac {a}{b}= \frac {\left(\frac {a}{h}\right)} {\left(\frac {b}{h}\right) }.</math>
They remain valid for angles greater than 90° and for negative angles.

:<math> \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} </math>

:<math> \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} </math>

:<math> \sec \theta = \frac {1}{\cos \theta} = \frac{\mathrm{hypotenuse}}{\mathrm{adjacent}} </math>
:<math> \csc \theta = \frac {1}{\sin \theta} = \frac{\mathrm{hypotenuse}}{\mathrm{opposite}} </math>

:<math> \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} </math>
Or
:<math> \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} </math>
:<math> \cot \theta = \frac {\csc \theta}{\sec \theta}</math>

===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&nbsp;−&nbsp;θ, obtaining:

:<math> \sin\left(  \pi/2-\theta\right) = \cos \theta</math>

:<math> \cos\left(  \pi/2-\theta\right) = \sin \theta</math>

:<math> \tan\left(  \pi/2-\theta\right) = \cot \theta</math>

:<math> \cot\left(  \pi/2-\theta\right) = \tan \theta</math>

:<math> \sec\left(  \pi/2-\theta\right) = \csc \theta</math>

:<math> \csc\left(  \pi/2-\theta\right) = \sec \theta</math>

===Pythagorean identities===
{{Main article|Pythagorean trigonometric identity}}
Identity 1:

:<math>\sin^2\theta + \cos^2\theta = 1</math>

The following two results follow from this and the ratio identities. To obtain the first, divide both sides of <math>\sin^2\theta + \cos^2\theta = 1</math> by <math>\cos^2\theta</math>; for the second, divide by <math>\sin^2\theta</math>.

:<math>\tan^2\theta + 1\ = \sec^2\theta </math>
:<math>\sec^2\theta - \tan^2\theta = 1 </math>

Similarly

:<math>1\ + \cot^2\theta = \csc^2\theta </math>
:<math>\csc^2\theta - \cot^2\theta = 1</math>

Identity 2:

The following accounts for all three reciprocal functions.

:<math> \csc^2\theta + \sec^2\theta - \cot^2\theta = 2\ + \tan^2\theta </math>

Proof 2:

Refer to the triangle diagram above. Note that <math>a^2+b^2=h^2</math> by [[Pythagorean theorem]].
:<math>\csc^2\theta + \sec^2\theta = \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}</math>

Substituting with appropriate functions -

:<math> 2\ + \frac{b^2}{a^2} + \frac{a^2}{b^2} = 2\ + \tan^2\theta+ \cot^2\theta </math>

Rearranging gives:

:<math> \csc^2\theta + \sec^2\theta - \cot^2\theta = 2\ + \tan^2\theta </math>

===Angle sum identities===
{{See also|List of trigonometric identities#Angle sum and difference identities}}

====Sine====
[[File:TrigSumFormula.svg|right|thumb|350px|Illustration of the sum formula.]]

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

Place P on the line defined by <math>\alpha + \beta</math> at a unit distance from the origin.

Let PQ be a line perpendicular to line OQ defined by angle <math>\alpha</math>, drawn from point Q on this line to point P. <math>\therefore</math> 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. <math>\therefore</math> OAQ and OBP are right angles.

Draw R on PB so that QR is parallel to the ''x''-axis.

Now angle <math>RPQ = \alpha</math> (because <math>OQA = \frac{\pi}{2} - \alpha</math>, making <math>RQO = \alpha, RQP = \frac{\pi}{2}-\alpha</math>, and finally <math>RPQ = \alpha</math>)
:<math>RPQ = \tfrac{\pi}{2} - RQP = \tfrac{\pi}{2} - (\tfrac{\pi}{2} - RQO) = RQO = \alpha</math>
:<math>OP = 1</math>
:<math>PQ = \sin \beta</math>
:<math>OQ = \cos \beta</math>
:<math>\frac{AQ}{OQ} = \sin \alpha</math>, so <math>AQ = \sin \alpha \cos \beta</math>
:<math>\frac{PR}{PQ} = \cos \alpha</math>, so <math>PR = \cos \alpha \sin \beta</math>
:<math>\sin (\alpha + \beta) = PB = RB+PR = AQ+PR = \sin \alpha \cos \beta + \cos \alpha \sin \beta</math>

By substituting <math>-\beta</math> for <math>\beta</math> and using the [[List of trigonometric identities#Reflections|reflection identities]] of [[even and odd functions]], we also get:
:<math>\sin (\alpha - \beta) = \sin \alpha \cos (-\beta) + \cos \alpha \sin (-\beta)</math>
:<math>\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta</math>

====Cosine====
Using the figure above,

:<math>OP = 1</math>

:<math>PQ = \sin \beta</math>

:<math>OQ = \cos \beta</math>

:<math>\frac{OA}{OQ} = \cos \alpha</math>, so <math>OA = \cos \alpha \cos \beta</math>

:<math>\frac{RQ}{PQ} = \sin \alpha</math>, so <math>RQ = \sin \alpha \sin \beta</math>

:<math>\cos (\alpha + \beta) = OB = OA-BA = OA-RQ = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta</math>

By substituting <math>-\beta</math> for <math>\beta</math> and using the [[List of trigonometric identities#Reflections|reflection identities]] of [[even and odd functions]], we also get:

:<math>\cos (\alpha - \beta) = \cos \alpha \cos (-\beta) - \sin \alpha \sin (-\beta),</math>
:<math>\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta</math>

Also, using the [[Complementary angle|complementary angle formulae]],

:<math>
\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}
</math>

====Tangent and cotangent====
From the sine and cosine formulae, we get
:<math>\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}</math>

Dividing both numerator and denominator by <math> \cos \alpha \cos \beta </math>, we get
:<math>\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}</math>

Subtracting <math> \beta </math> from <math> \alpha </math>, using <math>\tan (- \beta) = -\tan \beta </math>,
:<math>\tan (\alpha - \beta) = \frac{\tan \alpha + \tan (-\beta)}{1 - \tan \alpha \tan (-\beta)} = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}</math>

Similarly, from the sine and cosine formulae, we get
:<math>\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}</math>

Then by dividing both numerator and denominator by <math> \sin \alpha \sin \beta </math>, we get
:<math>\cot (\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}</math>

Or, using <math> \cot \theta = \frac{1}{\tan \theta} </math>,
:<math>\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} </math>

Using <math>\cot (- \beta) = -\cot \beta </math>,
:<math>\cot (\alpha - \beta) = \frac{\cot \alpha \cot (-\beta) - 1}{ \cot \alpha + \cot (-\beta) } = \frac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}</math>

===Double-angle identities===
From the angle sum identities, we get

:<math>\sin (2 \theta) = 2 \sin \theta \cos \theta</math>

and

:<math>\cos (2 \theta) = \cos^2 \theta - \sin^2 \theta</math>

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

:<math>\cos (2 \theta) = 2 \cos^2 \theta - 1</math>

:<math>\cos (2 \theta) = 1 - 2 \sin^2 \theta</math>

The angle sum identities also give

:<math>\tan (2 \theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2}{\cot \theta - \tan \theta}</math>

:<math>\cot (2 \theta) = \frac{\cot^2 \theta - 1}{2 \cot \theta} = \frac{\cot \theta - \tan \theta}{2}</math>

It can also be proved using [[Euler's formula]]

:<math> e^{i \varphi}=\cos \varphi +i \sin \varphi</math>

Squaring both sides yields

:<math> e^{i 2\varphi}=(\cos \varphi +i \sin \varphi)^{2}</math>

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

:<math> e^{i 2\varphi}=\cos 2\varphi +i \sin 2\varphi</math>

It follows that

:<math>(\cos \varphi +i \sin \varphi)^{2}=\cos 2\varphi +i \sin 2\varphi</math>.

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

:<math>i(2 \sin \varphi \cos \varphi) + \cos^2 \varphi - \sin^2 \varphi\ = \cos 2\varphi +i \sin 2\varphi</math>.

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

:<math>\cos^2 \varphi - \sin^2 \varphi\ = \cos 2\varphi</math>,

and also

:<math>2 \sin \varphi \cos \varphi = \sin 2\varphi</math>.

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

:<math>\cos \frac{\theta}{2} = \pm\, \sqrt\frac{1 + \cos \theta}{2},</math>

:<math>\sin \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{2}.</math>

The sign of the square root needs to be chosen properly—note that if 2{{pi}} 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:

:<math>\tan \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{1 + \cos \theta}.</math>

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

:<math>\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}.</math>

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

:<math>\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}.</math>

This also gives:

:<math>\tan \frac{\theta}{2} = \csc \theta - \cot \theta.</math>

Similar manipulations for the cot function give:

:<math>\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.</math>

===Miscellaneous – the triple tangent identity===

If <math>\psi + \theta + \phi = \pi = </math> half circle (for example, <math>\psi</math>, <math>\theta</math> and <math>\phi</math> are the angles of a triangle),
:<math>\tan(\psi) + \tan(\theta) + \tan(\phi) = \tan(\psi)\tan(\theta)\tan(\phi).</math>

Proof:<ref>{{cite web |url=http://mathlaoshi.com/tags/tangent-identity/ |title=Tangent Identity &#124; Math 老师 |accessdate=2013-10-30 |url-status=dead |archiveurl=https://web.archive.org/web/20131029203121/http://mathlaoshi.com/tags/tangent-identity/ |archivedate=2013-10-29 }} dead link</ref>
:<math>
\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}
</math>

===Miscellaneous – the triple cotangent identity===

If <math>\psi + \theta + \phi = \tfrac{\pi}{2} = </math> quarter circle,
:<math> \cot(\psi) + \cot(\theta) + \cot(\phi) = \cot(\psi)\cot(\theta)\cot(\phi)</math>.

Proof:

Replace each of <math>\psi </math>, <math>\theta </math>, and <math>\phi </math> with their complementary angles, so cotangents turn into tangents and vice versa.

Given
:<math>\psi + \theta + \phi = \tfrac{\pi}{2}</math>
:<math>\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 </math>

so the result follows from the triple tangent identity.

===Sum to product identities===

* <math>\sin \theta \pm \sin \phi = 2 \sin \left ( \frac{\theta\pm \phi}2 \right ) \cos \left ( \frac{\theta\mp \phi}2 \right )</math>
* <math>\cos \theta + \cos \phi = 2 \cos \left ( \frac{\theta+\phi}2 \right ) \cos \left ( \frac{\theta-\phi}2 \right ) </math>
* <math>\cos \theta - \cos \phi = -2 \sin \left ( \frac{\theta+\phi}2 \right ) \sin \left ( \frac{\theta-\phi}2 \right ) </math>

====Proof of sine identities====

First, start with the sum-angle identities:
:<math>\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta</math>
:<math>\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta</math>

By adding these together,
:<math>\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 </math>

Similarly, by subtracting the two sum-angle identities,
:<math>\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 </math>

Let <math>\alpha + \beta = \theta</math> and <math>\alpha - \beta = \phi</math>,
:<math>\therefore \alpha = \frac{\theta + \phi}2 </math> and <math>\beta = \frac{\theta - \phi}2 </math>

Substitute <math>\theta</math> and <math>\phi</math>
:<math>\sin \theta + \sin \phi = 2 \sin \left( \frac{\theta + \phi}2 \right) \cos \left( \frac{\theta - \phi}2 \right) </math>
:<math>\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) </math>

Therefore,
:<math>\sin \theta \pm \sin \phi = 2 \sin \left( \frac{\theta\pm \phi}2 \right) \cos \left( \frac{\theta\mp \phi}2 \right) </math>

====Proof of cosine identities====

Similarly for cosine, start with the sum-angle identities:
:<math>\cos (\alpha + \beta) = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta</math>
:<math>\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta</math>

Again, by adding and subtracting
:<math>\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</math>
:<math>\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</math>

Substitute <math>\theta</math> and <math>\phi</math> as before,
:<math>\cos \theta + \cos \phi = 2 \cos \left( \frac{\theta+\phi}2 \right) \cos \left( \frac{\theta-\phi}2 \right) </math>
:<math>\cos \theta - \cos \phi = -2 \sin \left( \frac{\theta+\phi}2 \right) \sin \left( \frac{\theta-\phi}2 \right) </math>

===Inequalities===
{{See also|List of triangle inequalities}}

[[File:TrigInequality.svg|right|thumb|342px|Illustration of the sine and tangent inequalities.]]

The figure at the right shows a sector of a circle with radius 1.  The sector is {{math|''θ''/(2{{pi}})}} of the whole circle, so its area is {{math|''θ''/2}}. We assume here that {{math|''θ'' < {{pi}}/2}}.

:<math>OA = OD = 1</math>
:<math>AB = \sin \theta</math>
:<math>CD = \tan \theta</math>

The area of triangle {{math|''OAD''}} is {{math|''AB''/2}}, or {{math|sin(''θ'')/2}}.  The area of triangle {{math|''OCD''}} is {{math|''CD''/2}}, or {{math|tan(''θ'')/2}}.

Since triangle {{math|''OAD''}} lies completely inside the sector, which in turn lies completely inside triangle {{math|''OCD''}}, we have

:<math>\sin \theta < \theta < \tan \theta.</math>

This geometric argument relies on definitions of [[arc length]] and
[[Jordan measure|area]], which act as assumptions, so it is rather a condition imposed in construction of [[trigonometric functions]] than
a provable property.<ref>
{{cite journal |last=Richman |first=Fred |title=A Circular Argument |journal=The College Mathematics Journal |date=March 1993 |volume=24 |issue=2 |pages=160–162 |jstor=2686787 |doi=10.2307/2686787}}</ref> For the sine function, we can handle other values.  If {{math|''θ'' > {{pi}}/2}}, then {{math|''θ'' > 1}}.  But {{math|sin ''θ'' ≤ 1}} (because of the Pythagorean identity), so {{math|sin ''θ'' < ''θ''}}.  So we have

:<math>\frac{\sin \theta}{\theta} < 1\ \ \ \mathrm{if}\ \ \ 0 < \theta.</math>

For negative values of {{math|''θ''}} we have, by the symmetry of the sine function

:<math>\frac{\sin \theta}{\theta} = \frac{\sin (-\theta)}{-\theta} < 1.</math>

Hence

:<math>\frac{\sin \theta}{\theta} < 1\quad \text{if }\quad \theta \ne 0,</math>
and
:<math>\frac{\tan \theta}{\theta} > 1\quad  \text{if }\quad  0 < \theta < \frac{\pi}{2}.</math>

==Identities involving calculus==

===Preliminaries===

:<math>\lim_{\theta \to 0}{\sin \theta} = 0</math>
:<math>\lim_{\theta \to 0}{\cos \theta} = 1</math>

===Sine and angle ratio identity===

:<math>\lim_{\theta \to 0}{\frac{\sin \theta}{\theta}} = 1</math>
In other words, the function sine is [[differentiable function|differentiable]] at 0, and its [[derivative]] is 1.

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

:<math>\sin \theta < \theta < \tan \theta</math>,
Therefore,
:<math>\frac{\sin \theta}{\theta} < 1 < \frac{\tan \theta}{\theta}</math>,
Consider the right-hand inequality. Since
:<math>\tan \theta = \frac{\sin \theta}{\cos \theta} </math>
:<math>\therefore 1 < \frac{\sin \theta}{\theta \cos \theta} </math>
Multiply through by <math>\cos \theta </math>
:<math>\cos \theta < \frac{\sin \theta}{\theta} </math>
Combining with the left-hand inequality:
:<math>\cos \theta < \frac{\sin \theta}{\theta} < 1 </math>
Taking <math>\cos \theta </math> to the limit as <math> \theta \to 0 </math>
:<math>\lim_{\theta \to 0}{\cos \theta} = 1</math>
Therefore,
:<math>\lim_{\theta \to 0}{\frac{\sin \theta}{\theta}} = 1</math>

===Cosine and angle ratio identity===
:<math>\lim_{\theta \to 0}\frac{1 - \cos \theta}{\theta} = 0</math>

Proof:

:<math>
\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}
</math>

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===
:<math> \lim_{\theta \to 0}\frac{1 - \cos \theta}{\theta^2}  = \frac{1}{2} </math>

Proof:

As in the preceding proof,

:<math>\frac{1 - \cos \theta}{\theta^2} = \frac{\sin \theta}{\theta} \times \frac{\sin \theta}{\theta} \times \frac{1}{1 + \cos \theta}.</math>

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
:<math>\sin[\arctan(x)]=\frac{x}{\sqrt{1+x^2}}</math>

Proof:

We start from
:<math>\sin^2\theta+\cos^2\theta=1</math> (I)
Then we divide this equation (I) by <math>\cos^2\theta</math>

:<math>\cos^2\theta=\frac{1}{\tan^2\theta+1}</math> (II)
:<math>1-\sin^2\theta=\frac{1}{\tan^2\theta+1}</math>
Then use the substitution <math>\theta=\arctan(x)</math>:

:<math>1-\sin^2[\arctan(x)]=\frac{1}{\tan^2[\arctan(x)]+1}</math>
:<math>\sin^2[\arctan(x)]=\frac{\tan^2[\arctan(x)]}{\tan^2[\arctan(x)]+1}</math>

Then we use the identity <math>\tan[\arctan(x)]\equiv x</math>
:<math>\sin[\arctan(x)]=\frac{x}{\sqrt{x^2+1}}</math> (III)

And initial Pythagorean trigonometric identity proofed...


Similarly if we divide this equation (I) by <math>\sin^2\theta</math>
:<math>\sin^2\theta=\frac{\frac{1}{1}}{1+\frac{1}{\tan^2\theta}}</math> (II)
:<math>\sin^2\theta=\frac{\tan^2\theta}{\tan^2\theta+1}</math>

Then use the substitution <math>\theta=\arctan(x)</math>:
:<math>\sin^2[\arctan(x)]=\frac{\tan^2[\arctan(x)]}{\tan^2[\arctan(x)]+1}</math>

Then we use the identity <math>\tan[\arctan(x)]\equiv x</math>
:<math>\sin[\arctan(x)]=\frac{x}{\sqrt{x^2+1}}</math> (III)

And initial Pythagorean trigonometric identity proofed...


:<math>[\arctan(x)]=[\arcsin(\frac{x}{\sqrt{x^2+1}})]</math>

:<math>y=\frac{x}{\sqrt{x^2+1}}</math>

:<math>y^2=\frac{x^2}{x^2+1}</math> (IV)


Let we guess that we have to prove:

:<math>x=\frac{y}{\sqrt{1-y^2}}</math>


:<math>x^2=\frac{y^2}{1-y^2}</math> (V)


Replacing (V) into (IV) :

:<math>y^2=\frac{\frac{y^2}{(1-y^2)}}{\frac{y^2}{(1-y^2)}+1}</math>

:<math>y^2=\frac{\frac{y^2}{(1-y^2)}}{\frac{1}{(1-y^2)}}</math>

So it's true: <math>y^2=y^2</math> and guessing statement was true: <math>x=\frac{y}{\sqrt{1-y^2}}</math>

:<math>[\arctan(x)]=[\arcsin(\frac{x}{\sqrt{x^2+1}})]=[\arcsin(y)]=[\arctan(\frac{y}{\sqrt{1-y^2}})]</math>


Now y can be written as x ; and we have [arcsin] expressed through [arctan]...


:<math>[\arcsin(x)]=[\arctan(\frac{x}{\sqrt{1-x^2}})]</math>


Similarly if we seek :<math>[\arccos(x)]</math>...

:<math>\cos[\arccos(x)]=x</math>

:<math>\cos(\frac{\pi}{2}-(\frac{\pi}{2}-[\arccos(x)]))=x</math>

:<math>\sin(\frac{\pi}{2}-[\arccos(x)])=x</math>

:<math>\frac{\pi}{2}-[\arccos(x)]=[\arcsin(x)]</math>

:<math>[\arccos(x)]=\frac{\pi}{2}-[\arcsin(x)]</math>


From :<math>[\arcsin(x)]</math>...

:<math>[\arccos(x)]=\frac{\pi}{2}-[\arctan(\frac{x}{\sqrt{1-x^2}})]</math>

:<math>[\arccos(x)]=\frac{\pi}{2}-[\arccot(\frac{\sqrt{1-x^2}}{x})]</math>

And finally we have [arccos] expressed through [arctan]...


:<math>[\arccos(x)]=[\arctan(\frac{\sqrt{1-x^2}}{x})]</math>

==See also==
{{div col|colwidth=22em}}
* [[List of trigonometric identities]]
* [[Bhaskara I's sine approximation formula]]
* [[Generating trigonometric tables]]
* [[Aryabhata's sine table]]
* [[Madhava's sine table]]
* [[Table of Newtonian series]]
* [[Madhava series]]
* [[Unit vector]] (explains direction cosines)
* [[Euler's formula]]
{{div col end}}

==Notes==
{{reflist}}

==References==
{{refbegin}}
* [[E. T. Whittaker]] and [[G. N. Watson]]. ''[[A Course of Modern Analysis]]'', [[Cambridge University Press]], 1952
{{refend}}

{{DEFAULTSORT:Trigonometric identities, Proofs of}}
[[Category:Trigonometry]]
[[Category:Article proofs]]