Find the values of the following:

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

Answers (1)

Here we have \cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )

let us assume that the value of

\cos^{-1}\left ( \frac{1}{2} \right ) = x, \:and\:\sin^{-1}\left(\frac{1}{2} \right ) = y;

then we have to find out the value of x +2y.

Calculation of x :

\Rightarrow \cos^{-1}\left ( \frac{1}{2} \right ) = x    

 \Rightarrow \cos x = \frac{1}{2}   

    \Rightarrow \cos \frac{\pi}{3} = \frac{1}{2}  , 

Hence x = \frac{\pi}{3}.

Calculation of y :

\Rightarrow \sin^{-1}\left(\frac{1}{2} \right ) = y  

 \Rightarrow \sin y = \frac{1}{2}    

   \Rightarrow \sin \frac{\pi}{6} = \frac{1}{2}  . 

  Hence y = \frac{\pi}{6}.

The required sum will be =     \frac{\pi}{3}+2(\frac{\pi}{6}) = \frac{2\pi}{3}.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions