त्रिकोणमितीय सर्वसमिकाओं की उपपत्तियाँ

इस लेख में प्रमुख त्रिकोणमितीय सर्वसमिकाओं की उपपतियां (सिद्धि) दी गयीं हैं।

## आरम्भिक त्रिकोणमितीय सर्वसमिकाएँ

### परिभाषाएँ

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}

### अनुपात वाली सर्वसमिकाएँ

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}

### 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:

: \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) = \sec \theta

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

### पाइथागोरीय सर्वसमिकाएँ
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) 

### कोणों के योग के त्रिकोणमितीय फलन

#### ज्या (Sine)

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

#### कोज्या (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}


#### स्पर्शज्या (Tangent) तथा कोस्पर्शज्या (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}

### दोगुने कोणों की सर्वसमिकाएँ

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.

### अर्ध कोणों की सर्वसमिकाएँ

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&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:

:\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.\,

### विविध

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:
:
\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) 

#### ज्या सर्वसमिकाओं की उपपत्ति

First, start with the sum-angle 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) 

#### कोज्या सर्वसमिकाओं की उपपत्ति

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 substracting
:\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) 

### असमिकाएँ

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 

This geometric argument applies if 0π/2, then θ>1. But sinθ≤1 (because of the Pythagorean identity), so sinθ\frac{\sin \theta}{\theta} 

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

:\frac{\sin \theta}{\theta} = \frac{\sin (-\theta)}{-\theta} 

Hence

:\frac{\sin \theta}{\theta} 

:\frac{\tan \theta}{\theta} > 1\ \ \ \mathrm{if}\ \ \ 0 

## कैलकुलस की सर्वसमिकाएँ

### प्रारम्भिक

:\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 ,
Therefore,
:\frac{\sin \theta}{\theta} ,
Consider the right-hand inequality. Since
:\tan \theta = \frac{\sin \theta}{\cos \theta} 
:\therefore 1 
Multply through by \cos \theta 
:\cos \theta 
Combining with the left-hand inequality:
:\cos \theta 
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}}

## इन्हें भी देखें
* त्रिकोणमितीय सर्वसमिकाओं की सूची

## सन्दर्भ
* E. T. Whittaker and G. N. Watson. *A course of modern analysis*, Cambridge University Press, 1952


श्रेणी:त्रिकोणमिति
श्रेणी:उपपत्ति