Mathematics Basics, Domain and Range , Concept and Need of Principal Brach, Graph Of Inverse Functions
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Basics, Domain and Range , Concept and Need of Principal Brach, Graph Of Inverse Functions

Basic Graphs of Trigonometry

1. `y = sinx `

Domain : ` -oo< x < oo`

Range : `-1 le y le 1`


2. `y = cosx`

Domain : `-oo < x< oo`

Range : `-1 le y le 1`


3. `y=tan x`

Domain : ` -oo< x < oo, x ne n pi + pi/2; n` `n in I`

Range : `-oo < y < oo`


4. `y= cosec (x)`

Domain : ` -oo< x < oo , x ne n pi; n in I`

Range : `y le -1` or `y ge 1`



5. `y= sec (x)`

Domain : ` -oo< x < oo , x ne n pi+pi/2; n in I`

Range : `y le -1` or `y ge 1`




6. `y= cot (x)`

Domain : ` -oo< x < oo, n ne n pi , n in I`

Range : `-oo < y < oo`

Existance of inverse of a function :

`=>` Inverse of a function of `f` exists, if the function is one-one and onto.

`=>` Since trigonometric functions are many-one over their domains, we restrict their domains and co-domains in order to make them one-one and onto and then find their inverse.

How to find principal branch of a trigonometric function :

I. Branch should cover whole range.

II. Branch should be closer to origin.

III. Branch should be continuous

IV. Prefer the branch in +ve domain to negative domain for principal branch.




1. `y = sinx `
Domain : `-pi/2 le x le pi/2`
Range : `-1 le y le 1`

`=> y = sin^(-1)x`
Domain : ` -1 le x le 1`
Range : `-pi/2 le y le pi/2`


2. `y = cosx`
Domain : `0 le x le pi`
Range : `-1 le y le 1`

`=> y = cos^(-1)x`
Domain : ` -1 le x le 1`
Range : `0 le y le pi`


3. `y=tan x`
Domain : `-pi/2 < x < pi/2`
Range : `-oo < y < oo`

`=> y = tan^(-1)x`
Domain : ` -oo < x < oo`
Range : `-pi/2 < y < pi/2`


4. `y= cosec (x)`
Domain : `-pi/2 le x le pi/2 , x ne 0`
Range : `y le -1` or `y ge 1`

`=> y = cosec^(-1)x`
Domain : ` x le -1` or `x ge 1`
Range : `-pi/2 le y le pi/2 , y ne 0`


5. `y= sec (x)`
Domain : `0 le x le pi , x ne pi/2`
Range : `y le -1` or `y ge 1`

`=> y = sec^(-1)x`
Domain : ` x le -1` or `x ge 1`
Range : `0 le y le pi , y ne pi/2`


6. `y= cot (x)`
Domain :`0 < x < pi`
Range : `-oo < y < oo`

`=> y = sec^(-1)x`
Domain : ` oo < x < oo`
Range : `0 < y< pi`
Q 2180545417

Find the principal values of `cos^(-1) (sqrt(3)/2)`
Class 12 Exercise 2.1 Q.No. 2
Solution:

Let `cos^(-1) (sqrt3/2) = theta`

`=> cos theta =sqrt3/2 = cos (pi/6)`

Therefore, principle value of `cos^(-1) (sqrt3/2)` is

`pi/6` as principal value of `cos^(-1) x` lies between `0` and `pi`
Q 2110645510

Find the principal values of `cos^(-1) (-1/2)`
Class 12 Exercise 2.1 Q.No. 5
Solution:

Let `cos^(-1) (-1/2) =theta`

`cos theta = -1/2 =cos (pi - pi/3) = cos\ (2pi)/3`

`:.` Principle values is `(2 pi)/3` as principle value of `cos^(-1) x` is `[0, pi]`
Q 2130645512

Find the principal values of `sec^(-1) (2/sqrt3)`
Class 12 Exercise 2.1 Q.No. 7
Solution:

Let `sec^(-1) (2/sqrt3) =y`

`=> sec y =2/sqrt3 = sec pi/6`

The range of principal value of `sec^(-1)` is

`[0, pi] - (pi/2)`


`:.` The principal value of `sec^(-1) (2/sqrt3)` is `pi/6`
Q 2160645515

Find the principal values of `cosec^(-1) ( -sqrt2)`
Class 12 Exercise 2.1 Q.No. 10
Solution:

Let `cosec^(-1) (- sqrt2) =y`

`=> cosec y = - sqrt 2 = - cosec \pi/4`


`=> cosec y = cosec (-pi/4)`

The range of principal value of `cosec^(-1)` is `[-pi/2 , pi/2] - {0}`

`:.` Principal value of `cosec^(-1) (- sqrt 2) = -pi/4`

Graph Of Inverse Function :

`"Method 1"` : The graph of an inverse trigonometric function can be obtained from the graph of original function by interchanging x-axis and y-axis, i.e, if (a, b) is a point on the graph of trigonometric function, then `(b, a)` becomes the corresponding point on the graph of
its inverse trigonometric function.

`"Method 2"` : It can be shown that the graph of an inverse function can be obtained from the corresponding graph of original function as a mirror image (i.e., reflection) along the line `y = x`.

Graph Of `y= sin^(-1) x` using method -1 :

- Draw principal branch of `y = sin x`





- Interchange `x` and `y` axis.






- Rotate the graph `90^o` anticlockwise.

Ploting of Graph of `sin^(-1) x` using mirror image about line `y=x`

`y=sin^(-1) x, |x| le 1, y in [-pi/2 , pi/2]`

`y= cos^(-1) x, |x| le 1 , y in [0, pi]`

`y= tan^(-1) x, x in R , y in (-pi/2 , pi/2)`

`y = cot^(-1) x, x in R, y in (0,pi)`

`y= cosec^-1 x, |x|ge1, y in [-pi/2,0)cup(0,pi/2]`

`y = sec^(-1)x , |x|ge 1, y in[0,pi/2)cup(pi/2,pi]`

Examples :

Q 2170645516

`tan^(-1) (1)+cos^(-1) (-1/2) + sin^(-1) (-1/2)`

Class 12 Exercise 2.1 Q.No. 11
Solution:

`tan^(-1 (1) + cos^(-1) (-1/2) + sin^(-1) (-1/2)`


Now `tan^(-1) =pi/4`

`:.` Range of principal value branch of `cos^(-1)` is `[0, pi]`

`cos^(-1) (-1/2) = pi - pi/3 =(2pi)/3`

and `sin^(-1) (-1/2) = - pi/6`


`:.` Range of principal value of `tan^(-1)` is `[-pi/2,pi/2]`

`:.` `tan^(-1) (1) + cos^(-1) (-1/2) + sin^(-1) (-1/2)`


`= pi/4 +(2 pi)/3 -pi/6 = (3 pi + 8pi -2pi)/12 = (9 pi)/12 =(3pi)/4`
Q 1881045827

`tan^(-1) sqrt(3) -sec^(-1) (-2)` is equal to
Class 12 Exercise 2.1 Q.No. 14
(A)

`pi`

(B)

`pi/3`

(C)

`pi/3`

(D)

`(2 pi)/3`

Solution:

`tan^(-1) sqrt(3) = pi/3` , `sec^(-1) (-2) = pi - pi/3 =(2 pi)/3`


`:.` Principal value of `sec^(-1)` is `[0- pi]- {pi/2}`


`:. tan^(-1) sqrt(3) - sec^(-1) (-2) =pi/3 - (2 pi)/3 =- pi/3`
Correct Answer is `=>` (A) `pi`
 
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