English

Sin { 2 Cos − 1 ( − 3 5 ) } is Equal to (A) 6 25 (B) 24 25 (C) 4 5 (D) − 24 25 - Mathematics

Advertisements
Advertisements

Question

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

 

Options

  • `6/25`

  • `24/25`

  • `4/5`

  • `-24/25`

MCQ

Solution

(d) `-24/25`

Let \[\cos^{- 1} \left( - \frac{3}{5} \right) = x, 0 \leq x \leq \pi\]
Then,
`cosx=-3/5`
\[\therefore \sin{x} = \sqrt{1 - \cos^2 x} = \sqrt{1 - \left( - \frac{3}{5} \right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}\]
Now,
\[\sin\left\{ 2 \cos^{- 1} \left( - \frac{3}{5} \right) \right\} = \sin\left( 2x \right)\]
\[ = 2\sin{x} \cos{x}\]
\[ = 2 \times \frac{4}{5} \times \frac{- 3}{5}\]
\[ = - \frac{24}{25}\]



shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.16 [Page 121]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 21 | Page 121

RELATED QUESTIONS

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


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


`sin^-1(sin  pi/6)`


Evaluate the following:

`sec^-1(sec  (7pi)/3)`


Evaluate the following:

`sec^-1{sec  (-(7pi)/3)}`


Evaluate the following:

`cosec^-1(cosec  (11pi)/6)`


Evaluate the following:

`cot^-1(cot  pi/3)`


Evaluate the following:

`cot^-1(cot  (4pi)/3)`


Write the following in the simplest form:

`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`


Evaluate the following:

`sec(sin^-1  12/13)`


Evaluate:

`tan{cos^-1(-7/25)}`


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x < 0


`sin^-1  5/13+cos^-1  3/5=tan^-1  63/16`


`sin^-1  4/5+2tan^-1  1/3=pi/2`


`2sin^-1  3/5-tan^-1  17/31=pi/4`


Prove that

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


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:

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


Write the difference between maximum and minimum values of  sin−1 x for x ∈ [− 1, 1].


If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,`  then write the value of x + y + z.


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


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


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


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


Write the principal value of \[\cos^{- 1} \left( \cos680^\circ  \right)\]


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


Find the value of \[\cos^{- 1} \left( \cos\frac{13\pi}{6} \right)\]


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

 


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


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

 

 


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

 


If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is

 


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


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


tanx is periodic with period ____________.


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


The value of tan `("cos"^-1  4/5 + "tan"^-1  2/3) =`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×