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:
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,
Or
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:
Identity 1:
Proof 1:
Refer to the triangle diagram above. Note that formula_22 by Pythagorean theorem.
The following two results follow from this and the ratio identities. To obtain the first, divide both sides of formula_24 by formula_25; for the second, divide by formula_26.
Similarly
Proof 2:
Differentiating the left-hand side of the identity yields:
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.
Proof 1:
Refer to the triangle diagram above. Note that formula_22 by Pythagorean theorem.
Substituting with appropriate functions -
Rearranging gives:
Draw a horizontal line (the "x"-axis); mark an origin O. Draw a line from O at an angle formula_37 above the horizontal line and a second line at an angle formula_38 above that; the angle between the second line and the "x"-axis is formula_39.
Place P on the line defined by formula_39 at a unit distance from the origin.
Let PQ be a line perpendicular to line defined by angle formula_37, drawn from point Q on this line to point P. formula_42 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. formula_42 OAQ and OBP are right angles.
Draw QR parallel to the "x"-axis.
Now angle formula_44 (because formula_45,  making formula_46, and finally formula_44)
By substituting formula_57 for formula_38 and using Symmetry, we also get:
Another simple "proof" can be given using Euler's formula known from complex analysis: Euler's formula is:
Although it is more precise to say that Euler's formula entails the trigonometric identities, it follows that for angles formula_37 and formula_38 we have:
Also using the following properties of exponential functions:
Evaluating the product:
Equating real and imaginary parts:
Using the figure above,
By substituting formula_57 for formula_38 and using Symmetry, we also get:
Also, using the complementary angle formulae,
From the sine and cosine formulae, we get
Dividing both numerator and denominator by formula_83, we get
Subtracting formula_85 from formula_86, using formula_87,
Similarly from the sine and cosine formulae, we get
Then by dividing both numerator and denominator by formula_90, we get
Or, using formula_92,
Using formula_94,
From the angle sum identities, we get
and
The Pythagorean identities give the two alternative forms for the latter of these:
The angle sum identities also give
It can also be proved using Euler's formula
Squaring both sides yields
But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields
It follows that
Expanding the square and simplifying on the left hand side of the equation gives
Because the imaginary and real parts have to be the same, we are left with the original identities
and also
The two identities giving the alternative forms for cos 2θ lead to the following equations:
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:
Then multiplying the numerator and denominator inside the square root by (1 + cos θ) and using Pythagorean identities leads to:
Also, if the numerator and denominator are both multiplied by (1 - cos θ), the result is:
This also gives:
Similar manipulations for the cot function give:
If formula_116, formula_117 and formula_118 are the angles of a triangle, i.e. formula_119 half circle,
Proof:
If formula_122 quarter circle,
Proof:
Replace each of formula_124, formula_125, and formula_126 with their complementary angles, so cotangents turn into tangents and vice-versa.
Given
so the result follows from the triple tangent identity.First, start with the sum-angle identities:
By adding these together,
Similarly, by subtracting the two sum-angle identities,
Let formula_136 and formula_137,
Substitute formula_117 and formula_118
Therefore,
Similarly for cosine, start with the sum-angle identities:
Again, by adding and substracting
Substitute formula_117 and formula_118 as before,
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.
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
This geometric argument applies if 0arc 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