Mathematics - Trigonometrical Ratios and Equations (6.11 - Proving of Identities)

The mathematics question is, prove the following identities:


Additional Mathematics
Picture is from http://www.singaporemath.com/PhotoGallery.asp?ProductCode=NSAMT

                                                           2
(n) (1 + cotӨ)² + (1 - cotӨ )² = ———
                                                       sin²Ө

Solution:

LHS = (1 + cotӨ)² + (1 - cotӨ )²

= 1 + 2cotӨ + cot²Ө + 1 - 2cotӨ + cot²Ө

= 2 + 2cot²Ө

               cos²Ө
= 2 + 2( ———— )
                sin²Ө

  2sin²Ө + 2cos²Ө
= ————————
           sin²Ө

     2(sin²Ө + cos²Ө)
= —————————
               sin²Ө

      2
= ———
   sin²Ө

= RHS (answer)



(o) tan²Ө - sin²Ө = tan²Ө sin²Ө

Solution:
LHS = tan²Ө - sin²Ө

    sin²Ө
= ———  -  sin²Ө
  cos²Ө

    sin²Ө - sin²Ө cos²Ө
= ——————————
               cos²Ө

     sin²Ө - (1 - cos²Ө)
= —————————
              cos²Ө

   sin²Ө sin²Ө
= ——————
          cos²Ө

= tan²Ө sin²Ө

= RHS (answer)



(q) (1 - cosx)(1 + secx) = sinx tanx


Solution:
LHS = (1 - cosx)(1 + secx)

                                  1
= (1 - cosx)( 1 + ——— )
                              cosx

                             1
= (1 - cosx) + ——— (1 - cosx)
                          cosx

                           1
= 1 - cosx + ———  - 1
                       cosx

     cosx - cos²x + 1 - cosx
= ————————————
                    cosx

    1 - cos²x
= —————
        cosx

    sin²x
= ———
    cosx

              sinx
= sinx ———
              cosx

= sinx tanx

= RHS (answer)





Standard Trigonometrical Ratios and Identities


              sinA                                             1
tanA = ———                     cosecA = ———                  sin²A + cos²A = 1
               cosA                                         sinA

               cosA                                       1
cotA = ———                     secA = ———                       1 + tan²A = sec²A
              sinA                                      cosA

                                                               1
                                              cotA = ———                        1 + cot²A = cosec²A
                                                            tanA


From "Additional Mathematics", 8th Edition, by The Keng Seng, Loh Cheng Yee, Consultant: Dr. Yeap Ban Har, Shinglee Publishers Pte Ltd, Chapter 6 - “Trigonometrical Ratios and Equations”, 6.11 - Proving of Identities, Page 180, table 6.1





Obtained from http://www.mathematics.com.au/images/600x480/5558.gif





Obtained from http://www.elcues.com/help/classX/XM8/XM8_Trigonometry_html_7c5771b3.png




Obtained from http://www.trigonometry-help.net/img/trig-tables_clip_image018.gif


Reference

• Additional Mathematics, 8th Edition, by The Keng Seng, Loh Cheng Yee, Consultant: Dr. Yeap Ban Har, Shinglee Publishers Pte Ltd, Chapter 6 - “Trigonometrical Ratios and Equations”, 6.11 - Proving of Identities, Page 180 - 183
• http://www.singaporemath.com/PhotoGallery.asp?ProductCode=NSAMT
• http://www.intmath.com/Analytic-trigonometry/1_Trigonometric-identities.php
• http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities
• http://www.mathematics.com.au/images/600x480/5558.gif
• http://www.elcues.com/help/classX/XM8/XM8_Trigonometry_html_7c5771b3.png http://www.trigonometry-help.net/img/trig-tables_clip_image018.gif - Black and White

Mathematics - Quadratic Equation (Part 1)

Picture is obtained from http://foldspace.files.wordpress.com/2008/06/c4743r.jpg

The mathematics question is,
Find the value of k of the quadratic equation

Such that one of its roots is thrice the value of the other.


Solution:
Let α and β be the roots. We have,

Sum of roots:


Product of roots:


Given that,



Substituting into eq. (1), we get



Therefore the other root is,


What to do next?
The roots, α and β are found but in terms of k.
How to find the value of k?
Note that equation (2) is not yet used.


Picture is obtained from http://plus.maths.org/issue29/features/quadratic/Quadratic.jpg

Substituting α and β into eq. (2), we have,




How to obtain the Quadratic Formula

From Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Quadratic_equation)
In mathematics, a quadratic equation is a polynomial equation of the second degree. The general form is



The quadratic coefficients are the following:
a is the coefficient of x2
b is the coefficient of x
c is the constant coefficient (also called the free term or constant term).

Quadratic equations are called quadratic because quadratus is Latin for "square".
The roots x or the quadratic formula can be derived by the method of completing the square. Dividing the quadratic equation



by a (which is allowed because a is non-zero), gives:



or



The quadratic equation is now in a form to which the method of completing the square can be applied.
To "complete the square" add the constant,

, to both sides,



The left side is now a perfect square because



Substituting the left side with the perfect square we have,



The right side can be written as a single fraction, with common denominator 4a2. This gives



Taking the square root of both sides yields



Isolating x, gives



This equation is known as the quadratic formula.




How to obtain Sum of roots and Product of roots

The general form is of a quadratic equation is


Dividing the quadratic equation by coefficient, a, we have


If a quadratic equation has roots, α and β, the equation can be written as


That is,



Comparing eq.(1) and eq.(2),




Reference

• http://en.wikipedia.org/wiki/Quadratic_equation
• Additional Mathematics, 8th Edition, by The Keng Seng, Loh Cheng Yee, Consultant: Dr. Yeap Ban Har, Shinglee Publishers Pte Ltd, Chapter 2 - “Quadratic Equations and Functions”, Page 32
• http://foldspace.files.wordpress.com/2008/06/c4743r.jpg (Blackboard)
• http://plus.maths.org/issue29/features/quadratic/Quadratic.jpg (Scary)
  
  

Mathematics - Indices and Algebra (Part 5)

The palm-like tree also joins the other fruit trees in blooming flowers. The surprise is that there are bees around. It must have honey to attract them.

The bees are busy and we are obviously being ignored. We also ignore them and carry on with our morning errand.

We cannot help it, but to take some more pictures. By this time there were many people crowding around. They are equally curious to know what is happening, with so many bees.

The mathematics question is,

Simplify


This question appears complex and quite intimidating.



Solution:
Let,


Substituting into the expression, we have,



Simplifying,





Summary of Index Laws

The laws are obtained from http://www.intmath.com/Numbers/exp.gif



Another representation of Index Rules

The Rules are obtained from http://www.revisioncentre.co.uk/gcse/maths/indices3.gif
Copyright © Matthew Pinkney 2003


Reference
http://www.intmath.com/Numbers/exp.gif
http://www.heymath.com/heymath/ (Login ID and Password required)
http://www.revisioncentre.co.uk/gcse/maths/indices3.gif

Mathematics - Indices and Algebra (Part 4) - Surds

Expressions with Indices, surds, exponentials and powers are the inverse of logarithm.
Indices and surds has real and known number.
Exponentials has unknowns, for example "X".
Logarithm of a number means the "exponential" of the base.
The flying pair of cranes is from http://www.cpedu.org/bbs/attachments/dvbbs/2006-6/200662918393174770.jpg

Normally in indices we come across square root of numbers giving results which are evaluated as exact number, just like the following.



However there are also square root of numbers that do not evaluate exactly. The examples below shows some of those irrational results also called irrational roots.






Numerical expressions which contain irrational roots are called surds.
Examples of surds. Conjugate surds exist too. They are a similar pair of expressions but one is with a negative and the other is with a positive irrational part.




Reference
http://www.heymath.com/heymath/ (requires login ID and password)
http://www.mathsteacher.com.au/year9/ch07_surds/03_surds/surds.htm
http://www.cpedu.org/bbs/attachments/dvbbs/2006-6/200662918393174770.jpg