त्रिकोणमितीय सर्वसमिकाओं की उपपत्तियाँ

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

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

${\displaystyle \sin \theta ={\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}={\frac {a}{h}}}$

${\displaystyle \cos \theta ={\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}={\frac {b}{h}}}$

${\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {a}{b}}}$

${\displaystyle \cot \theta ={\frac {\mathrm {adjacent} }{\mathrm {opposite} }}={\frac {b}{a}}}$

${\displaystyle \sec \theta ={\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}={\frac {h}{b}}}$

${\displaystyle \csc \theta ={\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}={\frac {h}{a}}}$

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

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

${\displaystyle \cot \theta ={\frac {\csc \theta }{\sec \theta }}}$

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:

${\displaystyle \sin \left(\pi /2-\theta \right)=\cos \theta }$

${\displaystyle \cos \left(\pi /2-\theta \right)=\sin \theta }$

${\displaystyle \tan \left(\pi /2-\theta \right)=\cot \theta }$

${\displaystyle \cot \left(\pi /2-\theta \right)=\tan \theta }$

${\displaystyle \sec \left(\pi /2-\theta \right)=\sec \theta }$

${\displaystyle \csc \left(\pi /2-\theta \right)=\csc \theta }$

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:

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

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:

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

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

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,

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

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

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

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

${\displaystyle 2\sin \varphi \cos \varphi =\sin 2\varphi }$.

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:

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

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:^[1]

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

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.

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

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,

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

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,

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

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

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.^[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

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

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

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

Hence

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

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

${\displaystyle \lim _{\theta \to 0}{\sin \theta }=0\,}$

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

${\displaystyle \lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}$

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,

${\displaystyle \lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}$

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

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.

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

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.

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 ${\displaystyle \tan[\arctan(x)]\equiv x}$

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

• त्रिकोणमितीय सर्वसमिकाओं की सूची

• 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.