त्रिकोणमितीय सर्वसमिकाओं की उपपत्तियाँ का स्रोत देखें

इस लेख में प्रमुख त्रिकोणमितीय [[सर्वसमिका]]ओं की उपपतियां (सिद्धि) दी गयीं हैं।

== आरम्भिक त्रिकोणमितीय सर्वसमिकाएँ==
{{अनुवाद}}
=== परिभाषाएँ===

[[Image:Trigonometric Triangle.svg|right|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.|396x396px]]

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

:<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>

=== अनुपात वाली सर्वसमिकाएँ===

The following identities are trivial algebraic consequences of these definitions and the division identity.<br>
They rely on multiplying or dividing the numerator and denominator of fractions by a variable. Ie,
: <math> \frac {a}{b}= \frac {\left(\frac {a}{c}\right)} {\left(\frac {b}{c}\right) }</math>

:<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>

=== पूरक कोणों से सम्बन्धित सर्वसमिकाएँ===
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;&minus;&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) = \sec \theta</math>

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

=== पाइथागोरीय सर्वसमिकाएँ===
Identity 1:

:<math>\sin^2(x) + \cos^2(x) = 1\,</math>

Proof 1:

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

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

:<math>\tan^2(x) + 1\ = \sec^2(x) </math>
:<math>\sec^2(x) - \tan^2(x) = 1\ </math>

Similarly

:<math> 1\ + \cot^2(x) = \csc^2(x) </math>
:<math>\csc^2(x) - \cot^2(x) = 1\ </math>

Proof 2:

Differentiating the left-hand side of the identity yields:

:<math>2 \sin x \cdot \cos x - 2 \sin x \cdot \cos x = 0 </math>

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.

:<math> \csc^2(x) + \sec^2(x) - \cot^2(x) = 2\ + \tan^2(x) </math>

Proof 1:

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

Substituting with appropriate functions -

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

Rearranging gives:

:<math> \csc^2(x) + \sec^2(x) - \cot^2(x) = 2\ + \tan^2(x) </math>

=== कोणों के योग के त्रिकोणमितीय फलन ===

==== ज्या (Sine)====
{{अनुवाद}}
[[Image: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 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 QR parallel to the ''x''-axis.

Now angle <math>RPQ = \alpha</math> (because <math>OQA = 90 - \alpha</math>, 
making <math>RQO = \alpha, RQP = 90-\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 [[List of trigonometric identities#Symmetry|Symmetry]], 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>

Another simple "proof" can be given using Euler's formula known from complex analysis:
Euler's formula is:
:<math>e^{i\varphi}=\cos \varphi +i \sin \varphi</math>

Although it is more precise to say that Euler's formula entails the trigonometric identities, it follows that for angles <math>\alpha</math> and <math>\beta</math> we have:
:<math>e^{i (\alpha + \beta)} = \cos (\alpha +\beta) + i \sin(\alpha +\beta)</math>

Also using the following properties of exponential functions:
:<math>e^{i(\alpha + \beta)} = e^{i \alpha} e^{i\beta}= (\cos \alpha +i \sin \alpha) (\cos \beta + i \sin \beta)</math>

Evaluating the product:
:<math>e^{i(\alpha + \beta)} = (\cos \alpha \cos \beta - \sin \alpha \sin \beta)+i(\sin \alpha \cos \beta + \sin \beta \cos \alpha)</math>

Equating real and imaginary parts:
:<math>\cos (\alpha +\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta</math>
:<math>\sin (\alpha +\beta)=\sin \alpha \cos \beta + \sin \beta \cos \alpha</math>

==== कोज्या (Cosine)====
{{अनुवाद}}
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 [[List of trigonometric identities#Symmetry|Symmetry]], 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 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>

==== स्पर्शज्या (Tangent) तथा कोस्पर्शज्या (cotangent)====
{{अनुवाद}}
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>

=== दोगुने कोणों की सर्वसमिकाएँ===
{{अनुवाद}}
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>.

=== अर्ध कोणों की सर्वसमिकाएँ===
{{अनुवाद}}
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&mdash;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:

:<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>

=== विविध ===

If <math>\psi</math>, <math>\theta</math> and <math>\phi</math> are the angles of a triangle, i.e. <math>\psi + \theta + \phi = \pi = </math> half circle,
:<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=संग्रहीत प्रति |access-date=8 जून 2014 |archive-url=https://web.archive.org/web/20131029203121/http://mathlaoshi.com/tags/tangent-identity/ |archive-date=29 अक्तूबर 2013 |url-status=dead }}</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.

=== Prosthaphaeresis 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>

==== ज्या सर्वसमिकाओं की उपपत्ति ====

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>

==== कोज्या सर्वसमिकाओं की उपपत्ति ====

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 substracting
:<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>

=== असमिकाएँ ===
[[Image: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 θ/(2π) of the whole circle, so its area is θ/2.

:<math>OA = OD = 1\,</math>

:<math>AB = \sin \theta\,</math>

:<math>CD = \tan \theta\,</math>

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

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

This geometric argument applies if 0<θ<π/2. It 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|url=http://www.jstor.org/stable/2686787 .|accessdate=3 नवम्बर 2012|doi=10.2307/2686787|archive-url=https://web.archive.org/web/20160317062314/http://www.jstor.org/stable/2686787|archive-date=17 मार्च 2016|url-status=live}}</ref> 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

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

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

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

Hence

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

:<math>\frac{\tan \theta}{\theta} > 1\ \ \ \mathrm{if}\ \ \ 0 < \theta < \frac{\pi}{2}\,</math>

== कैलकुलस की सर्वसमिकाएँ==

=== प्रारम्भिक ===

:<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>

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>
Multply 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> 
Then we divide this equation by <math>\cos^2\theta</math>
:<math>\cos^2\theta=\frac{1}{\tan^2\theta+1}</math>

Then use the substitution <math>\theta=\arctan(x)</math>, also use the Pythagorean trigonometric identity:
:<math>1-\sin^2[\arctan(x)]=\frac{1}{\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>

== इन्हें भी देखें ==
* [[त्रिकोणमितीय सर्वसमिकाओं की सूची]]

== सन्दर्भ ==
* E. T. Whittaker and G. N. Watson. ''A course of modern analysis'', Cambridge University Press, 1952
<references />

[[श्रेणी:त्रिकोणमिति]]
[[श्रेणी:उपपत्ति]]