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How Can We Draw The Graphs Of Trigonometric Functions?

published on December 3rd, 2006 . by Vanaja

We have four ways to alter the graph of a function: vertical translation, vertical scaling, horizontal translation, and horizontal scaling. Note that some or all four of these can be applied to a function at once in the sense that you may have to deal with a function of the form y = AsinB(x - C) + D. You should simply work these graphs out one step at a time, concerning yourself first with the affect of B and C, then the affect of the presence of A, and finally the affect of D.

Example:
Draw the graph of y =2 sin(x/2+?/6)+3

Solution
Let’s change this is of the form y = AsinB(x - C) + D
So y =2 sin(x/2+?/6)+3
can be written as y =2 sin ½ [x-(-?/3)]+3
Step 1: Draw sinx


B= ½ .so period is 4pi. That means it completes one cycle after 4pi.so stretch the graph horizontally so that its period is 4pi.

Next, we see that C=-pi/3. This will horizontally shift our graph pi/3units to the left. Thus, we now have


Now A=2, which will vertically scale our graph by a factor of 2. Thus, at this point, our range should now be [-2,2]

Finally, we must shift our graph vertically 3 units due to the presence of D=3. Hence, our range will now move to [1,5]

We can adopt the same method for drawing the graphs of other trigonometric functions. But remember first you should convert the given functions of the form

y = A sinB(x - C) + D

y = A cosB(x - C) + D

y = A tanB(x - C) + D

or accordingly, depending upon which of the trigonometric function is given.

Phase Shift Of Trigonometric Functions

published on November 27th, 2006 . by Vanaja

The phase shift is the horizontal shift away from the standard graph of the trigonometric function.

In y = AsinB(x - C) + D,C is the phase shift.
For cosine function y = AcosB(x - C) + D and tangent function y = AtanB(x - C) + D also C is the phase shift.

If the phase shift is positive, there has been a horizontal shift to the right and if it is negative, there has been a horizontal shift to the left.

In reading off the phase shift, make sure you have the function in the form above.

For example, the phase shift of y = sin(2x - ? /2) is not ?/2. Rewrite the expression for the function in the required form to get
y = sin2(x - ?/4).
Now we see the correct phase shift, is ?/4.


In the graph we can see graph of sin(2x - ? /2) is shifted to ?/4 units to the right.

Period Of Trigonometric Functions

published on November 26th, 2006 . by Vanaja

We know trigonometric functions are periodic functions. Sinx and cos x are periodic functions with period 2? or 360° . But tan and cot remain unchanged when x is increased by ? or 180° .
So, they are periodic functions with period ?.


The general form of the sine function is y = A sinB(x - C) + D
Here, the period is 2?/IBI.

The general form of the cosine function is y = A cosB(x - C) + D
We know cosine functions are identical to the sine functions. So, the period of cosine function is also 2?/IBI.
But for a tangent function. It is ? instead of 2? because the period of tan x is ?.
If the general form of a tangent function is y = A tanB(x - C)+ D,
its period is ?/IBI

There fore, the value of B is the key factor in determining the period of tangent functions. Change in its value changes horizontal stretching. When drawing the graph we have to “stretch” or ;“shrink” the graph horizontally by a factor of B.
Also, the period is unchanged by vertical scaling or shifting or by horizontal shifting.

Amplitude Of Trigonometric Functions

published on November 15th, 2006 . by Vanaja

We know basic trigonometric functions are sinx, cosx, tanx.
These functions are periodic functions.( The period is the shortest interval over which the function runs through one complete cycle of its graph.)
Sinx and cos x are periodic functions with period 2?.
But tan and cot remain unchanged when x is increased by pi.So, they are periodic functions with period ?.

Amplitude

See the graph of sinx . We know its range is [-1, 1].

It is clear from the graph that its amplitude is 1

When we draw the graph of 2 sin x, we can see that its range is [-2, 2]
The multiplication factor 2 has “stretched” the graph of sinx vertically by a factor of 2, while retaining the same x-intercepts.

This vertical scaling factor is known as the amplitude of the function.

The amplitude of the sine and cosine functions is half the vertical distance between its minimum value and its maximum value.

For a function A sin x, its y values range from –A to +A
So amplitude is 1/2 of [A-(-A) ]=A

The vertical shifts do alter the greatest and least values that the function attains but do not alter the amplitude.

We can verify this by taking the examples 2sin x and 2sinx+3 For 2sinx,the minimum and maximum values are -2 and 2 .

Amplitude is ½ . 2-(-2)=2
For 2sinx +3, minimum and maximum values are 1 and 5 .

Amplitude is ½ (5-1)=2

In general , the amplitude of


y = A sinB(x - C) + D and
y = A cosB(x - C) + D ,where B is a non-zero real number, is
IAI


The tangent function has no amplitude, because the tangent function has no minimum or maximum value.its range is (-infinity, infinity)

Area of a Triangle

published on October 11th, 2006 . by Vanaja

We know the formula for finding the area of a triangle is 1/2 bh, where b is the base and h is the height of the altitude from the opposite vertex to the base.

Another important formula is given below. We can use this formula when two sides and the included angle are given. We can derive Hero’s formula from this.

The area of triangle is given by

Similarly other results can also prove.

Projection Formulea

published on October 9th, 2006 . by Vanaja

In a triangle ABC,

a = b cosC + c cosB
b = c coaA + a cosC
c = a cosB + b cosA


These are known as projection formulea.

( To know the the parameters refer here. )


Proof
Let ABC be any of the triangles in the above figures

In fig (i) we have
a= BC = BD+DC……………..(a)
But BD/DA= cosB andDC/CA=cosC
==>BD=c cosB
and
DC=b cosC
Putting these values in (a)
a=c cosB+ bcosC

In fig(ii) BC=BD-CD ……………………… (b)
Here CD/CA=cos(180-C)=-cosC
==>CD=-bcosC
Also,BD=ccosB
Putting these values in (b)
a= c cosB-(-b cosC)
=c cosB+ b cosC

Other results can also be proved.

Similarly the results can also be proved for Fig(iii) also.

Example:

In any triangle ABC, prove that
(b +c )cos A+(c +a) cos B+ (a + b )cos C = a+ b+ c

Answer

Formula required: Projection Formula

In any triangle ABC, a = b cos C+c cos B

We have L.H.S=(b +c )cos A+(c +a) cos B+ (a + b )cos C

=b cos A+c cos A+c cos B+a cos B+ a cos C+ b cos C

=(b cos C+c cos B) + (a cos C+ c cos A) +(a cos B+b cos A) [ using projection formula]

=a+b+c

=R.H.S

History of Trigonometry

published on October 7th, 2006 . by Vanaja

The study of Trigonometry was first started in India. The ancient Indian mathematicians Aryabhatta (A.D 476),Bhaskara I(A.D 600), Bhaskara(A.D 1114)and Brahmagupta(A.D 598) got important results. All this knowledge first went from India to Middle East and from there to Europe. The Greeks had also started the study of trigonometry but their approach was so clumsy that when the Indian approach became known, it was immediately adopted throughout the world.
In India, the predecessor of the modern trigonometric function, known as the sine of an angle, and the introduction of the sine function represents the main contribution of the sidhantas to the history of mathematics.
Baskara I gave formula to find the values of sine function for angles more than 90 degree.
The name of Thales (A.D 600) is associated with height and distances problems. He is credited with the determination of the height of pyramid in Egypt by measuring shadows of the pyramid using similarity property.

The Law of Cosines

published on September 24th, 2006 . by Vanaja

The Law of Cosines

In any triangle ABC ,

If we know the three sides of any triangle, we can find the angles of the triangles using this formula.


Example:


In a triangle ABC, given a=25, b=52, c=63. Find A


Ans.

We have

Solutions of Triangle

published on September 20th, 2006 . by Vanaja

I think yesterday you had a big laugh after reading math genius. So today we are fresh and can have some lessons from Trigonometry.

We know the three sides and three angles of a triangle are called its parts. The process of finding the unknown parts of a triangle from its known parts is known as solution of a triangle. This has many applications in surveying, navigation, astronomy and other sciences.
In a Triangle ABC, The angles are denoted by A,B,C and the lengths of the corresponding opposite sides BC, CA and AB by a,b, and c respectively.

The Law of Sines


Using this formula we can find solutions of a triangle.If any three parts are given, we can find the other three parts.