English

`Sin^-1(Sin (7pi)/6)` - Mathematics

Advertisements
Advertisements

Question

`sin^-1(sin  (7pi)/6)`

Solution

We know

`sin(sin^-1theta)=theta if - pi/2<=theta<=pi/2`

We have

`sin^-1(sin  (7pi)/6)=sin^-1{sin(pi+pi/6)}`

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

`=-pi/6`

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.07 [Page 42]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.07 | Q 1.02 | Page 42

RELATED QUESTIONS

 

Prove that :

`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`

 

Solve the following for x:

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


If sin [cot−1 (x+1)] = cos(tan1x), then find x.


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


`sin^-1(sin12)`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`\text(cosec)^-1(\text{cosec}  pi/4)`


Evaluate the following:

`cot^-1{cot  ((21pi)/4)}`


Write the following in the simplest form:

`sin^-1{(sqrt(1+x)+sqrt(1-x))/2},0<x<1`


Evaluate:

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


Evaluate:

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


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x < 0


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x > 0


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


Solve the following equation for x:

tan−1(x + 1) + tan−1(x − 1) = tan−1`8/31`


Solve the following:

`cos^-1x+sin^-1  x/2=π/6`


`2sin^-1  3/5-tan^-1  17/31=pi/4`


`2tan^-1  1/5+tan^-1  1/8=tan^-1  4/7`


Show that `2tan^-1x+sin^-1  (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.


Solve the following equation for x:

`3sin^-1  (2x)/(1+x^2)-4cos^-1  (1-x^2)/(1+x^2)+2tan^-1  (2x)/(1-x^2)=pi/3`


Solve the following equation for x:

`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`


For any a, b, x, y > 0, prove that:

`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1  (2alphabeta)/(alpha^2-beta^2)`

`where  alpha =-ax+by, beta=bx+ay`


Write the value of sin1 (sin 1550°).


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


If \[\tan^{- 1} (\sqrt{3}) + \cot^{- 1} x = \frac{\pi}{2},\] find x.


What is the principal value of `sin^-1(-sqrt3/2)?`


Write the principal value of `tan^-1sqrt3+cot^-1sqrt3`


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


Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]


Write the value of  `cot^-1(-x)`  for all `x in R` in terms of `cot^-1(x)`


Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]


If  \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]


sin\[\left[ \cot^{- 1} \left\{ \tan\left( \cos^{- 1} x \right) \right\} \right]\]  is equal to

 

 

sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\]  is equal to

 


If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is

 


The value of  \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to

 


Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ;  1 < x < 1\].


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


Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`


The value of sin `["cos"^-1 (7/25)]` is ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×