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Lesson Posted on 03/01/2018 Learn Trigonometry

Trignometry

Raj Kumar

I am Six Sigma Black belt trained from American Society of Quality I am 2011 pass out in B.tech from...

Trignometry = Tri + Gon + Metern = (Three) + (Sides) + (Meausurment) = Measurement of three Sides/Angles read more

Trignometry = Tri         +  Gon     +   Metern

                    = (Three) +  (Sides) + (Meausurment)

                    = Measurement of three Sides/Angles

                                              Image result for perpendicular base hypotenuse

 

                            Image result for Pandit har har bole

                 http://slideplayer.com/slide/1468852/4/images/15/Some+People+Have+Curly+Brown+Hair+Tightly+Pulled+Back+Sin+Q+=+Perpendicular+Hypotenuse+Cos+Q+=+Base+Height+Tan+Q+=+Perpendicular+Base.jpg

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Lesson Posted on 09/06/2017 Learn Trigonometry

Unit Circle With Diameter = 1

Dhanushdhari

Unit 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... read more

                                                 Unit 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.

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Answered on 19/11/2016 Learn Trigonometry +2 Class XI-XII Tuition (PUC) Mathematics

Richa Kumari

1.7071
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Answered on 19/11/2016 Learn Trigonometry +2 Class XI-XII Tuition (PUC) Mathematics

Richa Kumari

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Answered on 19/11/2016 Learn Trigonometry +2 Class XI-XII Tuition (PUC) Mathematics

Richa Kumari

3?2 /2=2.1213
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Answered on 02/12/2016 Learn Trigonometry +2 Class XI-XII Tuition (PUC) Mathematics

Sarvajeet Kumar

An Experienced Trainer

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Answered on 03/12/2016 Learn Trigonometry +2 Class XI-XII Tuition (PUC) Mathematics

Gunturi Praveen

Maths & Electronics

Square the two equations and add. Apply the Pythagorean trigonometric Identity.
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Answered on 28/11/2016 Learn Trigonometry +2 Class XI-XII Tuition (PUC) Mathematics

Tapas Bhattacharya

Tutor

tan(x + y) = / = / = 1 = tan (pi/4) So, x + y = pi/4 radian = 45 degrees.
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Answered on 07/12/2016 Learn Trigonometry +2 Class XI-XII Tuition (PUC) Mathematics

Sreedevi Addepalli

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45 degrees.
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Answered on 03/12/2016 Learn Trigonometry +2 Class XI-XII Tuition (PUC) Mathematics

Sarvajeet Kumar

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