1. Angles,Measurement of Angles(Degree,Radian,Gradian) and Notational Convention
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Chapter 3
Trigonometric Functions
Angles, Measurement of Anges (Degree, Radian, Gradian) and Notational Convention:
Tri = three , gono= sides ,metric from metron(greek word) = study
The study of three sides is called trigonometry.
In other words , the trigonometry is that branch of mathematics which deals with measurement of sides and angles of a triangle.
Angles: a figure traced by rotating a given ray about its end points.
Otherwise An angle is formed when two straight lines or rays meet at a common endpoint. The common point of contact is called the vertex of an angle. The word angle comes from a Latin word named 'angulus,' meaning “corner.”
For measurement of sides, we need
- Units (cm,m,mm,hm,km,etc..)
- Devices ( Scale, tape, thread, gauge ,etc.,)
For measurement of angles , we need
- Units (degree, radian , gradian , staradian,etc..)
- Devices ( Protactor ,setsquare , angualar , trigonometric functions ,etc.,)
Three systems for measurement of angles.
- Sexagesimal system(FPS) or English System:
- Centesimal System(CGS) or French System
- Circular System
Sexagesimal system(FPS) or English System:
The sexagesimal system was an ancient system of counting, calculation, and numerical notation that used powers of 60 much as the decimal system uses powers of 10. Rudiments of the ancient system survive in vestigial form in our division of the hour into 60 minutes and the minute into 60 seconds.
In Sexagesimal System, an angle is measured in degrees, minutes and seconds.
A complete rotation describes 360°. In this system, a right angle is divided into 90 equal parts and each such part is called a Degree (1°); a degree is divided into 60 equal parts and each such part is called a Sexagesimal Minute (1’) and a minute is further sub-divided into 60 equal parts, each of which is called a Sexagesimal Second (1’’). In short,
Centesimal System(CGS) or French System:
In Centesimal System, an angle is measured in grades, minutes and seconds. In this system, a right angle is divided into 100
equal parts and each such part is called a Grade (1g); again, a grade is divided into 100 equal parts and each such part is called a Centesimal Minute (1‵) ; and a minute is further sub-divided into 100 equal parts, each of which is called a Centesimal Second (1‶). In short,
Note: (i) Clearly, minute and second in sexagesimal and centesimal systems are different.
For example,
The first relation is used to reduce an angle of sexagesimal system to centesimal system and the second is used to reduce an angle of centesimal system to sexagesimal system.
Circular System:
In this System, an angle is measured in radians. In higher mathematics angles are usually measured in circular system. In this system a radian is considered as the unit for the measurement of angles.
Definition of Radian: A radian is an angle subtended at the center of a circle by an arc whose length is equal to the radius.
The circular measure of an angle is the number of radians it contains.
Thus the circular (radian) measure of a right angle is
If an angle is given without mentioning units, it is assumed to be in radians. The relation between degree measures and circular (radian) measures of some standard angles are given below:
Radians = Degrees × π / 180
In terms of degrees, one complete counterclockwise revolution is 360° and whereas in radians, one complete counterclockwise revolution is 2π. These statements can be equated as:
One complete counterclockwise revolution in degrees = 360°
One complete counterclockwise revolution in radians = 2π
Radians = Degrees × (π/180°). We can follow the steps given below to calculate the measure of an angle given in degrees to radians.
- Note down the measure of the angle given in degrees.
- We know, 1°= (π)/180 radians. So, to convert the angle given in degrees to radians we multiply it with π/180°.
Angle in Radians = Angle in Degrees × π/180°. - Simplify the values and express the answer in radians.
Example: Convert 90 degrees to radians.
Solution: 90° = 90° × (π/180°) = π/2.
- Angles are measured in degrees, radians and gradian
- One full revolution is equal to 2π rad (or) 360°,100g.
- 1° = 0.017453 radians and 1 rad = 57.2958°.
- To convert an angle from degrees to radians, we multiply it by π/180°.
- To convert an angle from radians to degrees, we multiply it by 180°/π.
Example 1: In a circle with center O, points A and B are two points on the circle. ∠AOB = 60°. Convert ∠AOB's angle measure from degrees to radians.
Solution:
We have, ∠AOB = 60°.
Degrees to radians conversion formula is given as (Degrees × π)/180°
Thus converting the given angle we get,
∠AOB in Radians = ∠AOB in degrees × (π/180°)
∠AOB in Radians = 60° × (π/180°)
∠AOB in Radians = π/3
Answer: ∠AOB = π/3 rad.
Convert the following degree measure to radian measure.
a) 20 degrees b) 28 degrees
Solution: To convert the following degree measure to radians we will use the following steps:
a) 20° × (π/180°) = π/9 Radians.
b) 28° × (π/180°) = 7π/45 Radians.
2. Trigonometric Functions ,Sign,Domain ,Range and their graphs
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Trigonometric Functions, Sign, Domain, Range and their graphs
sin2x + cos2 x = 1
From the given identity, the following things can be interpreted:
cos2x = 1- sin2 x
cos x = √(1- sin2x)
Now we know that cosine function is defined for real values therefore the value inside the root is always non-negative. Therefore,
1- sin2x ≥ 0
sin2x ≤ 1
sin x ∈ [-1, 1]
Hence, we got the range and domain for sine function.
Similarly, following the same methodology,
1- cos2x ≥ 0
cos2x ≤1
cos x ∈ [-1,1]
Hence, for the trigonometric functions f(x)= sin x and f(x)= cos x, the domain will consist of the entire set of real numbers, as they are defined for all the real numbers. The range of f(x) = sin x and f(x)= cos x will lie from -1 to 1, including both -1 and +1, i.e.
- -1 ≤ sin x ≤1
- -1 ≤ cos x ≤1
Now, let us discuss the function f(x)= tan x. We know, tan x = sin x / cos x. It means that tan x will be defined for all values except the values that will make cos x = 0, because a fraction with denominator 0 is not defined. Now, we know that cos x is zero for the angles π/2, 3 π/2, 5 π/2 etc. therefore,
Hence, for these values, tan x is not defined.
So, the domain of f(x) = tan x will be R –
and the range will be set of all real numbers, R.
IMPORTANT NOTE
- The sum of interior angles of a polygon of n-sides = (n – 2) × 180º = (n – 2) π
- Each interior angle of a REGULAR polygon of n-sides =
- The sum of exterior angles of a polygon of any number of sides = 360º = 2π
BASIC FORMULAE OF TRIGONOMETRIC FUNCTIONS
SIGN OF TRIGONOMETRIC FUNCTIONS
Two perpendicular lines intersecting at a point O divide a plane in 4 right angles, each is called a quadrant. If θ be the angle that a line OA subtends with the initial line OX, anticlockwise then.
In I Quadrant, 0° < θ < 90°
In II Quadrant, 90° < θ < 180°
In III Quadrant, 180° < θ < 270°
In IV Quadrant, 270° < θ < 360°
The following table illustrates the sign of various trigonometric functions in all the four quadrants
PERIODIC FUNCTIONS
A function f is said to be periodic if there exists a real number
T > 0 such that f (x + T) = f (x) for all x. T is called the period of the function.
All trigonometric functions are periodic functions with following periods.
For sin θ, cos θ, sec θ, and cosec θ, the period = 2π. For
tan θ and cot θ the period = π
THEOREM : If f(x) is a periodic function with period T, then f (ax + b) is a periodic function with period
For example :
Period of sin θ is 2π
Period of sin 2θ is
Period of
Period of ....... etc.
PERIOD OF SOME IMPORTANT TRIGONOMETRIC FUNCTIONS
- Periods of sinnθ, cosnθ, secnθ, and cosecn θ = 2π if n is odd and π if n is even
- Periods of tannθ,cotnθ, = π, n even or odd.
- Periods of |sinθ|, |cosθ|, |tanθ|, |cot θ|, |secθ|, |cosec θ|= π
- Periods of |sin θ| + |cos θ|, |tan θ| + |cot θ|, |sec θ| + |cosec θ| are
Domain and Range for Sec, Cosec and Cot Functions
We know that sec x, cosec x and cot x are the reciprocal of cos x, sin x and tan x respectively. Thus,
sec x = 1/cos x
cosec x = 1/sin x
cot x = 1/tan x
Hence, these ratios will not be defined for the following:
- sec x will not be defined at the points where cos x is 0. Hence, the domain of sec x will be R-{(2n+1)π/2}, where n∈ Integer. The range of sec x will be R- (-1,1). Since, cos x lies between -1 to1, so sec x can never lie between that region.
- cosec x will not be defined at the points where sin x is 0. Hence, the domain of cosec x will be R-{nπ}, where n∈ Integer. The range of cosec x will be R- (-1,1). Since, sin x lies between -1 to1, so cosec x can never lie in the region of -1 and 1.
- cot x will not be defined at the points where tan x is 0. Hence, the domain of cot x will be R-{nπ}, where n∈ Integer. The range of cot x will be the set of all real numbers, R.
3. Trigonometric Functions of Sum and Difference of Two Angles
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Trigonometric Functions of Sum and Difference of Two Angles:
Consider the following figure:
A circle is drawn with center as origin and radius 1 unit. A point P1 is chosen at an angle of x units from x-axis. The co-ordinates are mentioned in the figure. Another point P2 is chosen, at an angle of y units from the line segment OP1. P3 is a point on the circle which is at an angle of y units from x-axis, measured clockwise.
Now, in the given figure, Δ OP1P3 is congruent to Δ OP2P4, by SAS congruency criteria.
Hence, P1P3 = P2P4 (CPCT)
⇒(P1P3 )2= (P2P4)2
Since we know the coordinates of all the four points, hence using distance formula, we can write:
[cos x – cos (-y)]2 + [sin x – sin (-y)]2 = [1- cos (x+y)]2 + sin2 (x+y)
On solving the above equation, we have the following identity:
Sum Formula for Cosine cos(x + y) = cos x cos y – sin x sin y ……… (1)
Replacing y by -y in identity (1), we get,
cos(x – y) = cos x cos y + sin x sin y …….… (2)
Also,
cos (π/2 – x) = sin x ……………… (3)
That can be obtained by replacing x by π/2 and y by x in identity (2). Also,
sin (π/2 – x) = cos x………….…… (4)
As, sin (π/2 – x) = cos [π/2 – (π/2 – x)] (using identity 3). So,
sin (π/2 – x) = cos x
Now we have the idea about the expansion of sum and difference of angles of cos. Now let us try to use it for finding the values of sum and difference of angles of sin.
sin (x + y) can be written as cos [π/2 – (x + y)] which is equal to cos [(π/2 – x) – y]
Now, using identity (2) we can write,
cos [(π/2 – x) – y] = cos (π/2 – x) cos y + sin (π/2 – x) sin y
= sin x cos y + cos x sin y
Hence,
Sum Formula for Sine sin (x + y) = sin x cos y + cos x sin y …………………………. (5)
Replace y by –y in the above formula, we get
sin (x – y) = sin x cos y – cos x sin y .…………………………….. (6)
Now if we substitute suitable values in identities (1), (2), (5) and (6), we have the following:
cos (π/2 + x) = -sin x
sin (π/2 + x) = cos x
cos (π± x) = – cos x
sin (π – x) = sin x
sin (π + x) = – sin x
sin (2π – x) = -sin x
cos (2π – x) = cos x
After having a brief idea about the expansion of sum and difference of angles of sin and cos, the expansion for tan and cot is given by
tan (x + y) = (tan x + tan y)/ (1-tan x tan y)
tan (x – y) = (tan x – tan y)/ (1+tan x tan y)
Similarly;
cot (x + y) = (cot x cot y – 1)/(cot y + cot x)
cot (x – y) = (cot x cot y + 1)/(cot y – cot x)
EXTRA:
Generalisation :
- sin (A1 + A2 + ...... + An) = cos A1 cos A2 ...... cos An (S1 – S3 + S5 – ....)
- cos (A1 + A2 + ...... + An) = cos A1 cos A2 ...... cos An (1 – S2 + S4 – S6 + ....)
- tan (A1 + A2 + ...... + An) =
where S1 = Σ tan A , S2 = Σ tan A1 tan A2,
S3 = Σ tan A1 tan A2 tan A3 and so on.
FORMULAS TO TRANSFORM PRODUCTS INTO SUM AND DIFFERENCE
- 2 sin A cos B = sin(A+B) + sin (A – B)
- 2 cos A sin B = sin (A+B) – sin (A – B)
- 2 cos A cos B = cos (A+B) + cos (A – B)
- 2 sin A sin B = cos (A – B) – cos (A + B)
FORMULAS TO TRANSFORM SUM OR DIFFERENCE INTO PRODUCT
TRIGONOMETRIC FUNCTIONS OF MULTIPLE AND SUB-MULTIPLE ANGLES
FORMULAS FOR LOWERING THE DEGREE OF TRIGONOMETRIC FUNCTIONS
CONDITIONAL TRIGONOMETRIC IDENTITIES
If A + B + C = 180° (or π), or A, B, C are angles of a triangle. Then
4. Trigonometric Equations
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Trigonometric Equations:
Equations having trigonometric functions of a variable is known as Trigonometric Equations.
The equation is not completely solved unless we obtain an expression for all angles which satisfy it.
Example: sin2x + 7 cos x – 7 = 0 ,
sin 3x + 3 cos2x = 7 , etc.
All trigonometric ratios are periodic in nature, a trigonometric equation has in general, an infinite no. of solution.
So , the solutions of trigonometric equation may be categorized as:
Types of Solutions:
- General solution
- Principal solution
Principal Solution :
The solutions of these equations for a trigonometric function in variable x, where x lies in between 0≤x≤2π is called as principal solution.
General Solution:
If the solution contains the integer ‘n’ ,which gives all solution, it is called as the general solution.
Sometimes:
Particular Solution:
A Specific Value of x (the unknown angle) which satisfy the given equation.
N.B.: 1. The General Solution. contains Principal Solution as well as Particular solution
2. Principal Solution always two
Table
Proof :
sin x = sin y implies x = nπ + (-1)n y , where n € Z and x and y are any real numbers.
Q 1: If f(x) = tan 3x, g(x) = cot (x – 60) and h(x) = cos x,
find x given f(x) = g(x). Also, if h(x) = 4/5, find cosec x + tan3x.
Solution: If f(x) = g(x), so tan 3x = cot (x – 50)
=>cot (90 – 3x) = cot (x-50)
=> 90 – 3x = x – 50
or x = 35
For h(x)=cos x and h(x) = 4/5, we have cos x = 4/5.
Therefore, sin x = 3/5, cosec x = 5/3 and tan x = 4/5
Or, cosec x + tan3x = (5/3) + (4/5)3 = 817/375 = 2.178
Q 2: Find the principal solutions of the equation tan x = – 1/(√3).
Solution: We know, tan(π/6) = 1/(√3)
Since, tan (π – π/6 ) = -tan(π/6) = – 1/(√3)
Further, tan (2π – π/6) = -tan(π/6) = – 1/(√3)
Hence, the principal solutions are tan (π – π/6) = tan (5π/6) and tan (2π – π/6 ) = tan (11π/6)
Q 3: Evaluate the value of sin (11π/12).
Solution:
sin (11π/12) can be written as sin (2π/3 + π/4)
using formula, sin (x + y) = sin x cos y + cos x sin y
sin (11π/12) = sin (2π/3 + π/4) = sin(2π/3) cos (π/4) + cos(2π/3) sin (π/4)
= (√3)/2 x √2/2 + (-1/2) x √2/2
= √6/4 – (√2)/4
= (√6-√2)/4
Q 4: Evaluate cosec x = 2.
Solution: We know, cosec x = cosec π/6 = 2 or sin x = sin π/6 = 1/2 .
- x = n π + (-1)n π/6
Q 5: Solve 5 cos2y + 2 sin y = 0.
Solution: 5 cos2y + 2 sin y = 0
- 5 (1 – sin2 y) + 2 sin y = 0
- 5 sin2y – 2 sin y – 5 =0
- sin y = 1.2 or sin y = -0.8.
Since sin y can not be greater than 1,
sin y = – 0.8 = sin ( π + π/3 )
or, sin y = sin (4π/3) and hence, the solution is given by y = n π + (-1)n (4π/3)
Q 6: Find the principal solutions of the equation sin x = (√3/2)
Solution: Let y=sin-1 (√3/2)
- sin y = (√3/2)
- sin y= sin( π/3) and sin (2π/3) = sin (π – π/3 ) = sin π/3 = (√3/2)
Therefore, the principal solutions are x = π/3 and 2π/3.