English

Write the Difference Between Maximum and Minimum Values Of Sin−1 X For X ∈ [− 1, 1]. - Mathematics

Advertisements
Advertisements

Question

Write the difference between maximum and minimum values of  sin−1 x for x ∈ [− 1, 1].

Advertisements

Solution

The maximum value of `sin^-1x` in x ∈ [− 1, 1] is at 1.
So, the maximum value is

`sin^-1(1)`

`=sin^-1(sin  pi/2)`

`=pi/2`

Again, the minimum value is at -1.
Thus, the minimum value is

`sin^-1(-1)=-sin^-1(1)`

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

`=-pi/2`

So, the difference between the maximum and the minimum value is `pi/2-(-pi/2)=pi`

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.15 [Page 116]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 2 | Page 116

RELATED QUESTIONS

 

Show that:

`2 sin^-1 (3/5)-tan^-1 (17/31)=pi/4`

 

 

If (tan1x)2 + (cot−1x)2 = 5π2/8, then find x.


Prove that

`tan^(-1) [(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]=pi/4-1/2 cos^(-1)x,-1/sqrt2<=x<=1`


`sin^-1(sin4)`


Evaluate the following:

`cos^-1(cos12)`


Evaluate the following:

`tan^-1(tan2)`


Evaluate the following:

`sec^-1(sec  (25pi)/6)`


Evaluate the following:

`cot^-1{cot  ((21pi)/4)}`


Write the following in the simplest form:

`tan^-1{(sqrt(1+x^2)+1)/x},x !=0`


Evaluate the following:

`sin(tan^-1  24/7)`


Evaluate the following:

`cos(tan^-1  24/7)`


Evaluate:

`sec{cot^-1(-5/12)}`


Evaluate:

`cosec{cot^-1(-12/5)}`


Evaluate:

`cos(tan^-1  3/4)`


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x > 0


`sin^-1x=pi/6+cos^-1x`


`4sin^-1x=pi-cos^-1x`


`tan^-1x+2cot^-1x=(2x)/3`


Solve the following equation for x:

`tan^-1  2x+tan^-1  3x = npi+(3pi)/4`


`sin^-1  5/13+cos^-1  3/5=tan^-1  63/16`


If `cos^-1  x/2+cos^-1  y/3=alpha,` then prove that  `9x^2-12xy cosa+4y^2=36sin^2a.`


Evaluate the following:

`sin(1/2cos^-1  4/5)`


`tan^-1  2/3=1/2tan^-1  12/5`


`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`


Solve the following equation for x:

`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`


Prove that:

`tan^-1  (2ab)/(a^2-b^2)+tan^-1  (2xy)/(x^2-y^2)=tan^-1  (2alphabeta)/(alpha^2-beta^2),`   where `alpha=ax-by  and  beta=ay+bx.`


If −1 < x < 0, then write the value of `sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))`


Write the value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]


Write the value of sin1 (sin 1550°).


Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]


Write the value of \[\sin^{- 1} \left( \frac{1}{3} \right) - \cos^{- 1} \left( - \frac{1}{3} \right)\]


Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]


If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.

 

2 tan−1 {cosec (tan−1 x) − tan (cot1 x)} is equal to


If 4 cos−1 x + sin−1 x = π, then the value of x is

 


If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 .\]


The period of the function f(x) = tan3x is ____________.


Find the value of `sin^-1(cos((33π)/5))`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×