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

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


If `(sin^-1x)^2 + (sin^-1y)^2+(sin^-1z)^2=3/4pi^2,`  find the value of x2 + y2 + z2 


`sin^-1{(sin - (17pi)/8)}`


`sin^-1(sin3)`


Evaluate the following:

`cos^-1{cos(-pi/4)}`


Evaluate the following:

`cos^-1(cos4)`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`sec^-1(sec  (9pi)/5)`


Evaluate the following:

`cot^-1(cot  pi/3)`


Evaluate the following:

`cot^-1(cot  (19pi)/6)`


Evaluate the following:

`cosec(cos^-1  3/5)`


Evaluate:

`cos(tan^-1  3/4)`


If `cos^-1x + cos^-1y =pi/4,`  find the value of `sin^-1x+sin^-1y`


Solve the following equation for x:

tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x


Solve the following equation for x:

 cot−1x − cot−1(x + 2) =`pi/12`, > 0


Solve the following equation for x:

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


Evaluate: `cos(sin^-1  3/5+sin^-1  5/13)`


`2tan^-1  3/4-tan^-1  17/31=pi/4`


Prove that

`sin{tan^-1  (1-x^2)/(2x)+cos^-1  (1-x^2)/(2x)}=1`


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


Find the value of the following:

`cos(sec^-1x+\text(cosec)^-1x),` | x | ≥ 1


Solve the following equation for x:

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


Write the value of

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


Write the value of cos1 (cos 350°) − sin−1 (sin 350°)


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


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


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


The number of real solutions of the equation \[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x), - \pi \leq x \leq \pi\]


\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]


\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\]  is equal to

 

 


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

 


\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\] 

 


The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]

 


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


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


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×