0Unit Circle For Advanced Trigonometry
Important: Please draw the circles and lines as directed as you read along.
In modern school syllabi for teaching trigonometric functions, a unit circle is defined as a circle with radius equal to one unit. The radius is the hypotenuse of the right angled triangle. Probably this is also the international practice.
However, in the erstwhile Hyderabad state of Nizam, before 1947 and for a few years afterwards till the syllabi were adopted according to the practice in rest of India, for the entrance examination of the Osmania University to explain the trigonometric functions the unit circle was defined as a circle with diameter (not radius) equal to one.
The medium of instruction in the schools was Urdu which used the Hindi terms derived from Sanskrit. A brief overview of the topic is given below.
Draw a circle with AB as the diameter of the circle. So the length of AB is one unit. => AB = 1
Take a point C on the circumference. Angle BCA is the right angle.
BC is the chord of angle A (i.e. angle BAC).
Please remember that angle A (i.e. BAC) is the same where ever A may be on the arc.
The chord BC was called Jya of the angle A Sine of A *
And chord AC was called Kotijya of angle A Cosine of A
Draw the tangent at B and extend AC to meet it at D
Tangent BD was called Sparsh of angle A Tangent of A
Draw the tangent at A and extend BC to meet it at E
Tangent AE was called Kotisparsh of angle A Cotangent of A
Secant AD was called Chhedak of angle A Secant of A
And secant BE was called Kotichhedak of A Cosecant of A
* It was said that Jya was mispronounced by the Arabs as Sya which was modified by the Europeans to Sine. The other terms are simple translations
Sine (A+B)
A similar approach was used to get the sine and cosine functions of the sum of two angles.
Draw PQRS, a cyclic quadrilateral with diagonal PR as the diameter = 1.
= > triangles PQR and PSR are right angled with common hypotenuse PR = 1.
Let QPR be angle A and RPS be angle B.
= > sin A = QR, cos A = PQ, sin B = SR, cos B = PS and SQ = sin (A+B).
From the geometry of a cyclic quadrilateral with one diagonal as the diameter,
PR x QS = PQ x RS + QR x PS.
Since PR = 1, we get sin (A+B) = sin A x cos B + cos A x sin B.
Cos (A+B)
Now draw a chord ST at right angles to SQ, cos (A+B) = ST.
Join PT and QT
Since triangles PQR and PQT are congruent, PT = QR = sin A.
In the cyclic quadrilateral PQST, QT x PS = PQ x ST + PT x QS.
= > Since QT = PR = 1, cos B = cos A x cos (A+B) + sin A x sin (A+B).
The theorem can be proved by expanding sin (A+B) as proved earlier.
Cos (A-B)
Similarly For cos (A-B), Let angle QPS = A, and angle QPR = B.
=> angle RQS = A-B and SP = cos (A-B),
Cos A = ST, cos B = PQ, sin A = SQ and sin B = QR
In the cyclic quadrilateral PQST, QT x PS = PQ x ST + PT x QS
Since QT = 1, we get cos (A-B) = cos A x cos B + sin A x sin B
The theorem for sin (A-B) can also be proved similarly.
---0---
