Class 11 Maths Trigonometric Functions Domain & Range

Domain & Range of Trigonometric Functions

 

Using the table we can observe that Sin & Cos are defined for all real numbers. Further, we observe that for each real number x, – 1 ≤ sin x ≤ 1 and – 1 ≤ cos x ≤ 1.

Thus, domain of y = sin x and y = cos x is the set of all real numbers and range is the interval [–1, 1], i.e., – 1 ≤ y ≤ 1.

 

Graph of Sin x & Cos x is shown. Both repeats after 2π

 

 

If we notice the graph of tan x, ire repeats after π. Thus domain of y = tan x is the set {x : x ∈ R and x ≠ (2n + 1) π/2 , n ∈ Z} and range is the set of all real numbers.

Note that x ≠ (2n + 1) π/2 is not part of domain since for these values of x, tan x is undefined.

 

 

Similarly domain & range of cosec, sec & cot can be defined.

 

  • Since cosec x = 1/sin x , the domain of y = cosec x is the set { x : x ∈ R and x ≠ n π, n ∈ Z} and range is the set {y : y ∈ R, y ≥ 1 or y ≤ – 1}.

 

  • The domain of y = sec x is the set {x : x ∈ R and x ≠ (2n + 1) π/2 , n ∈ Z} and range is the set {y : y ∈ R, y ≤ – 1or y ≥ 1}.

 

  • The domain of y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real numbers.

 

 

 

 Numerical: If cos x = – 3/ 5 , x lies in the third quadrant, find the values of other five trigonometric fx

Solution: Cos x = Base/ Hypotenuse = -3/5

p2 = h2 – b2          (using Pythagoras Theorem)

or p2 = 52 – 32   

or  p = 4

Now, it is given that x lies in 3rd quadrant, so sin & cos will be negative & tan will be positive.

So sin x = -4/5  , cos x =  - 3/5   & tan x = 4/3

 

 

Numerical: Find the value of sin(31π/3) &  cos (–1710°)

Solution: Sin (31π/3) = sin (10 π  +  π/3)  = sin (π/3) = √3/2

 

Cos (–1710°) = cos (–1710° + 5 * 360o)   = cos ( -1710o + 1800o)  = cos(90o)  = 0

 

Trigonometric Functions

  • sin (- x) = - sin x
  • cos (- x) = cos x
  • tan(-x) = - tan x
  • cosec(-x) = -cosec x
  • sec (-x) = sec x
  • cot (-x) = -cot x

 

  • cos (π/2-x) = sin x
  • sin (π/2 –x )= cos x
  • tan (π/2 –x )= cot x
  • cot (π/2 –x )= tan x

 

  • cos (π/2+ x ) =  - sin x
  • sin (π/2 +x) = cos x
  • tan (π/2+ x ) =  - cot x
  • cot (π/2 +x) = - tan x

 

  • cos (π – x) = - cos x
  • sin (π – x) = sin x
  • tan(π – x) = -tan x

 

  • cos (π + x) = - cos x
  • sin (π + x) =  -sin x
  • tan(π + x) = tan x

 

  • sin (x + y) = sin x cos y + cos x sin y
  • cos (x + y) = cos x cos y – sin x sin y
  • tan (x + y) =(tan x +tan y)/ 1 –tan x * tan y

 

  • sin (x – y) = sin x cos y – cos x sin y
  • cos (x – y) = cos x cos y + sin x sin y
  • tan (x - y) =(tan x -tan y)/ 1 +tan x * tan y

 

  • cos 2x = cos2x – sin2 x
  • sin 2x = 2 sinx cos x
  • tan 2x=  2tan x / (1- tan2 x)

 

  • sin 3x = 3 sin x – 4 sin3 x
  • cos 3x = 4 cos3 x – 3 cos x
  • tan 3x = (3tan x – tan3 x)/(1-3 tan2 x)
  • 1 + tan2 x = sec2 x
  • 1 + cot2 x = cosec2
  • 1- sin2 x = cos2 x

 

  • cos x cos y = (cos (x + y) + cos (x – y))/2
  • sin x sin y = (cos (x - y) – cos (x + y))/2
  • sin x cos y =( sin (x + y) + sin (x – y))/2
  • cos x sin y = (sin (x + y) – sin (x – y))/2

 

  • cos x + cos y = 2 cos (x+ y)/2 cos (x – y)/2
  • cos x – cos y = – 2sin (x+ y)/2  sin (x- y)/2
  • sin x + sin y = 2sin (x+ y)/2  cos(x- y)/2
  • sin x – sin y = 2cos (x+ y)/2  sin (x- y)/2 

 

 

Numerical: Find the value of sin 15°.

 

Solution:

 sin 15° =sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30°  = 1/√2 * √3/2  - 1/√2 * ½   = (√3 -1)/2√2

 

Numerical: Prove that       sin(x+y) / sin(x-y)   = ( tan x + tan y)/ tan x - tan y)

 

Solution: 

sin(x+y) / sin(x-y)    =   (sin x * cos y +  cos x *  sin y) /(sin x * cos y -  cos x *  sin y)

 

Divide numerator & denominator by cos x cos y, to get

sin(x+y) / sin(x-y)   = ( tan x + tan y)/ tan x - tan y)

 

For more numerical, refer ExamFear video lessons.

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