त्रिकोणमितीय सर्वसमिकाओं की उपपत्तियाँ

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:

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

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

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

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

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

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

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 − 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)

Illustration of the sum formula.

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,

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

स्पर्शज्या (Tangent) तथा कोस्पर्शज्या (cotangent)

From the sine and cosine formulae, we get

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

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

$e^{i2\varphi }=(\cos \varphi +i\sin \varphi )^{2}$

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

$e^{i2\varphi }=\cos 2\varphi +i\sin 2\varphi$

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—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:

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

Miscellaneous -- the triple cotangent identity

If $\psi +\theta +\phi ={\tfrac {\pi }{2}}=$ 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

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.

$OA=OD=1\,$

$AB=\sin \theta \,$

$CD=\tan \theta \,$

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

$\sin \theta <\theta <\tan \theta \,$

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

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

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

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

Hence

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

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

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

$\sin \theta <\theta <\tan \theta \,$,

Therefore,

${\frac {\sin \theta }{\theta }}<1<{\frac {\tan \theta }{\theta }}\,$,

Consider the right-hand inequality. Since

$\tan \theta ={\frac {\sin \theta }{\cos \theta }}$

$\therefore 1<{\frac {\sin \theta }{\theta \cos \theta }}$

Multply through by $\cos \theta$

$\cos \theta <{\frac {\sin \theta }{\theta }}$

Combining with the left-hand inequality:

$\cos \theta <{\frac {\sin \theta }{\theta }}<1$

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:

${\begin{aligned}{\frac {1-\cos \theta }{\theta }}&={\frac {1-\cos ^{2}\theta }{\theta (1+\cos \theta )}}\\&={\frac {\sin ^{2}\theta }{\theta (1+\cos \theta )}}\\&=\left({\frac {\sin \theta }{\theta }}\right)\times \sin \theta \times \left({\frac {1}{1+\cos \theta }}\right)\\\end{aligned}}$

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

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

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