हिंदी

Evaluate the Following: `Sin(1/2cos^-1 4/5)` - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following:

`sin(1/2cos^-1  4/5)`

उत्तर

`sin(1/2cos^-1  4/5)=sin{1/2xx2sin^-1+-sqrt((1-4/5)/2)}`       `[becausecos^-1x=2sin^-1+-sqrt((1-x)/2)]`

`=sin(sin^-1+-1/sqrt10)`

`=+-1/sqrt10`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.14 [पृष्ठ ११५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.14 | Q 1.3 | पृष्ठ ११५

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Solve the equation for x:sin1x+sin1(1x)=cos1x


Solve the following for x:

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


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`


​Find the principal values of the following:

`cos^-1(-1/sqrt2)`


`sin^-1(sin3)`


Evaluate the following:

`cos^-1{cos  ((4pi)/3)}`


Evaluate the following:

`tan^-1(tan  (6pi)/7)`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`sec^-1(sec  (7pi)/3)`


Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`


Write the following in the simplest form:

`tan^-1sqrt((a-x)/(a+x)),-a<x<a`


Evaluate the following:

`sin(sec^-1  17/8)`


Evaluate: 

`cot(sin^-1  3/4+sec^-1  4/3)`


Find the value of `tan^-1  (x/y)-tan^-1((x-y)/(x+y))`


Solve the following equation for x:

`tan^-1  x/2+tan^-1  x/3=pi/4, 0<x<sqrt6`


Solve the following equation for x:

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


`tan^-1  1/7+2tan^-1  1/3=pi/4`


If `sin^-1  (2a)/(1+a^2)-cos^-1  (1-b^2)/(1+b^2)=tan^-1  (2x)/(1-x^2)`, then prove that `x=(a-b)/(1+ab)`


Find the value of the following:

`tan^-1{2cos(2sin^-1  1/2)}`


Solve the following equation for x:

`tan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`


Solve the following equation for x:

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


If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.


Write the range of tan−1 x.


Write the value of cos\[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]


Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]


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


Write the value of \[\cos\left( \sin^{- 1} x + \cos^{- 1} x \right), \left| x \right| \leq 1\]


The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is

 


If sin−1 − cos−1 x = `pi/6` , then x = 


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

 


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 =

 


The domain of  \[\cos^{- 1} \left( x^2 - 4 \right)\] is

 


If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When  \[\theta = \frac{\pi}{3}\] .


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 \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.


Find the domain of `sec^(-1)(3x-1)`.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×