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

आरम्भिक त्रिकोणमितीय सर्वसमिकाएँ
--------------------------------

### परिभाषाएँ

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:

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

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

{\displaystyle \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 }}}
{\displaystyle \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 − θ, obtaining:

{\displaystyle \sin \left(\pi /2-\theta \right)=\cos \theta }
{\displaystyle \cos \left(\pi /2-\theta \right)=\sin \theta }
{\displaystyle \tan \left(\pi /2-\theta \right)=\cot \theta }
{\displaystyle \cot \left(\pi /2-\theta \right)=\tan \theta }
{\displaystyle \sec \left(\pi /2-\theta \right)=\sec \theta }
{\displaystyle \csc \left(\pi /2-\theta \right)=\csc \theta }
### पाइथागोरीय सर्वसमिकाएँ

Identity 1:

{\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1\,}
Proof 1:

Refer to the triangle diagram above. Note that {\displaystyle a^{2}+b^{2}=h^{2}} by Pythagorean theorem.

{\displaystyle \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 {\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1} by {\displaystyle \cos ^{2}(x)}; for the second, divide by {\displaystyle \sin ^{2}(x)}.

{\displaystyle \tan ^{2}(x)+1\ =\sec ^{2}(x)}
{\displaystyle \sec ^{2}(x)-\tan ^{2}(x)=1\ }
Similarly

{\displaystyle 1\ +\cot ^{2}(x)=\csc ^{2}(x)}
{\displaystyle \csc ^{2}(x)-\cot ^{2}(x)=1\ }
Proof 2:

Differentiating the left-hand side of the identity yields:

{\displaystyle 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.

{\displaystyle \csc ^{2}(x)+\sec ^{2}(x)-\cot ^{2}(x)=2\ +\tan ^{2}(x)}
Proof 1:

Refer to the triangle diagram above. Note that {\displaystyle a^{2}+b^{2}=h^{2}} by Pythagorean theorem.

{\displaystyle \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 -

{\displaystyle 2\ +{\frac {b^{2}}{a^{2}}}+{\frac {a^{2}}{b^{2}}}=2\ +\tan ^{2}(x)+\cot ^{2}(x)}
Rearranging gives:

{\displaystyle \csc ^{2}(x)+\sec ^{2}(x)-\cot ^{2}(x)=2\ +\tan ^{2}(x)}
### कोणों के योग के त्रिकोणमितीय फलन

#### ज्या (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 {\displaystyle \alpha } above the horizontal line and a second line at an angle {\displaystyle \beta } above that; the angle between the second line and the *x*-axis is {\displaystyle \alpha +\beta }.

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

Let PQ be a line perpendicular to line defined by angle {\displaystyle \alpha }, drawn from point Q on this line to point P. {\displaystyle \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. {\displaystyle \therefore } OAQ and OBP are right angles.

Draw QR parallel to the *x*-axis.

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

{\displaystyle RPQ={\tfrac {\pi }{2}}-RQP={\tfrac {\pi }{2}}-({\tfrac {\pi }{2}}-RQO)=RQO=\alpha }
{\displaystyle OP=1}
{\displaystyle PQ=\sin \beta }
{\displaystyle OQ=\cos \beta }
{\displaystyle {\frac {AQ}{OQ}}=\sin \alpha \,}, so {\displaystyle AQ=\sin \alpha \cos \beta }
{\displaystyle {\frac {PR}{PQ}}=\cos \alpha \,}, so {\displaystyle PR=\cos \alpha \sin \beta }
{\displaystyle \sin(\alpha +\beta )=PB=RB+PR=AQ+PR=\sin \alpha \cos \beta +\cos \alpha \sin \beta }
By substituting {\displaystyle -\beta } for {\displaystyle \beta } and using Symmetry, we also get:

{\displaystyle \sin(\alpha -\beta )=\sin \alpha \cos -\beta +\cos \alpha \sin -\beta }
{\displaystyle \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:

{\displaystyle 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 {\displaystyle \alpha } and {\displaystyle \beta } we have:

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

{\displaystyle e^{i(\alpha +\beta )}=e^{i\alpha }e^{i\beta }=(\cos \alpha +i\sin \alpha )(\cos \beta +i\sin \beta )}
Evaluating the product:

{\displaystyle 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:

{\displaystyle \cos(\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta }
{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\sin \beta \cos \alpha }
#### कोज्या (Cosine)

Using the figure above,

{\displaystyle OP=1\,}
{\displaystyle PQ=\sin \beta \,}
{\displaystyle OQ=\cos \beta \,}
{\displaystyle {\frac {OA}{OQ}}=\cos \alpha \,}, so {\displaystyle OA=\cos \alpha \cos \beta \,}
{\displaystyle {\frac {RQ}{PQ}}=\sin \alpha \,}, so {\displaystyle RQ=\sin \alpha \sin \beta \,}
{\displaystyle \cos(\alpha +\beta )=OB=OA-BA=OA-RQ=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta \,}
By substituting {\displaystyle -\beta } for {\displaystyle \beta } and using Symmetry, we also get:

{\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos -\beta \ -\sin \alpha \sin -\beta \,}
{\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta \,}
Also, using the complementary angle formulae,

{\displaystyle {\begin{aligned}\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{aligned}}}
#### स्पर्शज्या (Tangent) तथा कोस्पर्शज्या (cotangent)

From the sine and cosine formulae, we get

{\displaystyle \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 {\displaystyle \cos \alpha \cos \beta }, we get

{\displaystyle \tan(\alpha +\beta )={\frac {\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}}
Subtracting {\displaystyle \beta } from {\displaystyle \alpha }, using {\displaystyle \tan(-\beta )=-\tan \beta },

{\displaystyle \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

{\displaystyle \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 {\displaystyle \sin \alpha \sin \beta }, we get

{\displaystyle \cot(\alpha +\beta )={\frac {\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }}}
Or, using {\displaystyle \cot \theta ={\frac {1}{\tan \theta }}},

{\displaystyle \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 {\displaystyle \cot(-\beta )=-\cot \beta },

{\displaystyle \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

{\displaystyle \sin(2\theta )=2\sin \theta \cos \theta \,}
and

{\displaystyle \cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta \,}
The Pythagorean identities give the two alternative forms for the latter of these:

{\displaystyle \cos(2\theta )=2\cos ^{2}\theta -1\,}
{\displaystyle \cos(2\theta )=1-2\sin ^{2}\theta \,}
The angle sum identities also give

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

{\displaystyle e^{i\varphi }=\cos \varphi +i\sin \varphi }
Squaring both sides yields

{\displaystyle e^{i2\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

{\displaystyle e^{i2\varphi }=\cos 2\varphi +i\sin 2\varphi }
It follows that

{\displaystyle (\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

{\displaystyle 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

{\displaystyle \cos ^{2}\varphi -\sin ^{2}\varphi \ =\cos 2\varphi },
and also

{\displaystyle 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:

{\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}},\,}
{\displaystyle \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:

{\displaystyle \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:

{\displaystyle \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:

{\displaystyle \tan {\frac {\theta }{2}}={\frac {1-\cos \theta }{\sin \theta }}.\,}
This also gives:

{\displaystyle \tan {\frac {\theta }{2}}=\csc \theta -\cot \theta .\,}
Similar manipulations for the cot function give:

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

{\displaystyle \tan(\psi )+\tan(\theta )+\tan(\phi )=\tan(\psi )\tan(\theta )\tan(\phi ).}
Proof:

{\displaystyle {\begin{aligned}\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{aligned}}}
### Miscellaneous -- the triple cotangent identity

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

{\displaystyle \cot(\psi )+\cot(\theta )+\cot(\phi )=\cot(\psi )\cot(\theta )\cot(\phi )}.
Proof:

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

Given

{\displaystyle \psi +\theta +\phi ={\tfrac {\pi }{2}}\,}
{\displaystyle \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

* {\displaystyle \sin \theta \pm \sin \phi =2\sin \left({\frac {\theta \pm \phi }{2}}\right)\cos \left({\frac {\theta \mp \phi }{2}}\right)}
* {\displaystyle \cos \theta +\cos \phi =2\cos \left({\frac {\theta +\phi }{2}}\right)\cos \left({\frac {\theta -\phi }{2}}\right)}
* {\displaystyle \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:

{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }
{\displaystyle \sin(\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta }
By adding these together,

{\displaystyle \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,

{\displaystyle \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 {\displaystyle \alpha +\beta =\theta } and {\displaystyle \alpha -\beta =\phi },

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

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

{\displaystyle \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:

{\displaystyle \cos(\alpha +\beta )=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta }
{\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta }
Again, by adding and substracting

{\displaystyle \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 \ }
{\displaystyle \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 {\displaystyle \theta } and {\displaystyle \phi } as before,

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

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.

{\displaystyle OA=OD=1\,}
{\displaystyle AB=\sin \theta \,}
{\displaystyle 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

{\displaystyle \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. 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

{\displaystyle {\frac {\sin \theta }{\theta }}<1\ \ \ \mathrm {if} \ \ \ 0<\theta \,}
For negative values of θ we have, by symmetry of the sine function

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

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

कैलकुलस की सर्वसमिकाएँ
----------------------

### प्रारम्भिक

{\displaystyle \lim \_{\theta \to 0}{\sin \theta }=0\,}
{\displaystyle \lim \_{\theta \to 0}{\cos \theta }=1\,}
### Sine and angle ratio identity

{\displaystyle \lim \_{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}
Proof: From the previous inequalities, we have, for small angles

{\displaystyle \sin \theta <\theta <\tan \theta \,},
Therefore,

{\displaystyle {\frac {\sin \theta }{\theta }}<1<{\frac {\tan \theta }{\theta }}\,},
Consider the right-hand inequality. Since

{\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}}
{\displaystyle \therefore 1<{\frac {\sin \theta }{\theta \cos \theta }}}
Multply through by {\displaystyle \cos \theta }

{\displaystyle \cos \theta <{\frac {\sin \theta }{\theta }}}
Combining with the left-hand inequality:

{\displaystyle \cos \theta <{\frac {\sin \theta }{\theta }}<1}
Taking {\displaystyle \cos \theta } to the limit as {\displaystyle \theta \to 0}

{\displaystyle \lim \_{\theta \to 0}{\cos \theta }=1\,}
Therefore,

{\displaystyle \lim \_{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}
### Cosine and angle ratio identity

{\displaystyle \lim \_{\theta \to 0}{\frac {1-\cos \theta }{\theta }}=0}
Proof:

{\displaystyle {\begin{aligned}{\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{aligned}}}
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

{\displaystyle \lim \_{\theta \to 0}{\frac {1-\cos \theta }{\theta ^{2}}}={\frac {1}{2}}}
Proof:

As in the preceding proof,

{\displaystyle {\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 

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

We start from

{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
Then we divide this equation by {\displaystyle \cos ^{2}\theta }

{\displaystyle \cos ^{2}\theta ={\frac {1}{\tan ^{2}\theta +1}}}
Then use the substitution {\displaystyle \theta =\arctan(x)}, also use the Pythagorean trigonometric identity:

{\displaystyle 1-\sin ^{2}[\arctan(x)]={\frac {1}{\tan ^{2}[\arctan(x)]+1}}}
Then we use the identity {\displaystyle \tan[\arctan(x)]\equiv x}

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

इन्हें भी देखें
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* त्रिकोणमितीय सर्वसमिकाओं की सूची

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

1. ↑  "संग्रहीत प्रति". मूल से 29 अक्तूबर 2013 को पुरालेखित. अभिगमन तिथि 8 जून 2014.
2. ↑  Richman, Fred (March 1993).. "A Circular Argument" जाँचें `|url=` मान (मदद). *The College Mathematics Journal*. **24** (2): 160–162. डीओआइ:10.2307/2686787. मूल से 17 मार्च 2016 को पुरालेखित. अभिगमन तिथि 3 नवम्बर 2012.