मराठी

Evaluate Sin \[\Left( \Frac{1}{2} \Cos^{- 1} \Frac{4}{5} \Right)\] - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate sin

\[\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right)\]

Advertisements

उत्तर

We know that

\[\cos^{- 1} x = 2 \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}\]
\[ \tan^{- 1} x = \sin^{- 1} \frac{x}{\sqrt{1 + x^2}}\]

\[\therefore \sin\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right) = \sin\left( \frac{1}{2}2 \tan^{- 1} \sqrt{\frac{1 - \frac{4}{5}}{1 + \frac{4}{5}}} \right)\]
\[ = \sin\left( \tan^{- 1} \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}} \right)\]
\[ = \sin\left( \tan^{- 1} \frac{1}{3} \right)\]
\[ = \sin\left\{ \sin^{- 1} \left( \frac{\frac{1}{3}}{\sqrt{1 + \frac{1}{9}}} \right) \right\}\]
\[ = \sin\left( \sin^{- 1} \frac{1}{\sqrt{10}} \right)\]
\[ = \frac{1}{\sqrt{10}} \left[ \because \sin\left( \sin^{- 1} x \right) = x \right]\]

∴ \[\sin\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right) = \frac{1}{\sqrt{10}}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.15 [पृष्ठ ११७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 17 | पृष्ठ ११७

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

 

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`


 

Show that:

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

 

 

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


Evaluate the following:

`cos^-1{cos  ((4pi)/3)}`


Evaluate the following:

`tan^-1(tan1)`


Evaluate the following:

`sec^-1(sec  (5pi)/4)`


Evaluate the following:

`cot^-1(cot  (4pi)/3)`


Evaluate the following:

`cot^-1{cot  ((21pi)/4)}`


Write the following in the simplest form:

`cot^-1  a/sqrt(x^2-a^2),|  x  | > a`


Write the following in the simplest form:

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


Prove the following result

`cos(sin^-1  3/5+cot^-1  3/2)=6/(5sqrt13)`


Evaluate:

`cot{sec^-1(-13/5)}`


Evaluate:

`cos(tan^-1  3/4)`


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


Solve the following equation for x:

`tan^-1(2+x)+tan^-1(2-x)=tan^-1  2/3, where  x< -sqrt3 or, x>sqrt3`


Sum the following series:

`tan^-1  1/3+tan^-1  2/9+tan^-1  4/33+...+tan^-1  (2^(n-1))/(1+2^(2n-1))`


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


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)`


Solve the following equation for x:

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


Write the value of sin−1

\[\left( \sin( -{600}°) \right)\].

 

 


Write the value of cos−1 \[\left( \tan\frac{3\pi}{4} \right)\]


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).


Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]


Write the principal value of `sin^-1(-1/2)`


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


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


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


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

 


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

 

 

Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) = 


The value of \[\cos^{- 1} \left( \cos\frac{5\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{5\pi}{3} \right)\] is

 


If \[3\sin^{- 1} \left( \frac{2x}{1 + x^2} \right) - 4 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) = \frac{\pi}{3}\] is equal to

 


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×