English

Evaluate the Following: `Cos^-1(Cos12)` - Mathematics

Advertisements
Advertisements

Question

Evaluate the following:

`cos^-1(cos12)`

Solution

We know

`cos^-1(costheta)=thetaif 0<=theta<=pi`

We have

`cos^-1(cos12)=cos^-1{cos(4pi-12)}`

= 4π - 12

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 2.8 | Page 42

RELATED QUESTIONS

Find the domain of definition of `f(x)=cos^-1(x^2-4)`


Evaluate the following:

`cos^-1{cos  ((4pi)/3)}`


Evaluate the following:

`tan^-1(tan2)`


Evaluate the following:

`cot^-1(cot  pi/3)`


Evaluate the following:

`cot^-1(cot  (19pi)/6)`


Evaluate the following:

`cot^-1{cot (-(8pi)/3)}`


Write the following in the simplest form:

`tan^-1{x+sqrt(1+x^2)},x in R `


Evaluate the following:

`sin(sec^-1  17/8)`


Evaluate the following:

`cos(tan^-1  24/7)`


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


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


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


`sin^-1  63/65=sin^-1  5/13+cos^-1  3/5`


Solve the following:

`cos^-1x+sin^-1  x/2=π/6`


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


Solve the following equation for x:

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


Write the value of tan1x + tan−1 `(1/x)`for x > 0.


Write the value of

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


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


If tan−1 x + tan−1 y = `pi/4`,  then write the value of x + y + xy.


Write the value of cos−1 (cos 6).


Write the value of \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right)\]


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


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


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

 

 

If tan−1 3 + tan−1 x = tan−1 8, then x =


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

 


It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\]   (−7), then the value of x is

 


The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is

 


If tan−1 (cot θ) = 2 θ, then θ =

 


If > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to

 


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


If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×