+91-8076753736[email protected]
logo
logo
-->

Inverse Trigonometric Functions

Question
CBSEENMA12032780
Wired Faculty App

Prove cos to the power of negative 1 end exponent space x equals 2 space sin to the power of negative 1 end exponent space square root of fraction numerator 1 minus x over denominator 2 end fraction end root equals 2 space cos to the power of negative 1 end exponent square root of fraction numerator 1 plus x over denominator 2 end fraction end root

Solution

Put x = cos straight theta
therefore space space space space cos to the power of negative 1 end exponent straight x equals cos to the power of negative 1 end exponent left parenthesis cos space straight theta right parenthesis equals straight theta space space space space space space Also space 2 space sin to the power of negative 1 end exponent space square root of fraction numerator 1 minus straight x over denominator 2 end fraction end root equals 2 space sin to the power of negative 1 end exponent square root of fraction numerator 1 minus cosθ over denominator 2 end fraction end root space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space equals 2 space sin to the power of negative 1 end exponent square root of fraction numerator open parentheses 2 space sin squared begin display style straight theta over 2 end style close parentheses over denominator 2 end fraction end root equals 2 space sin to the power of negative 1 end exponent open parentheses sin straight theta over 2 close parentheses equals 2 open parentheses straight theta over 2 close parentheses equals straight theta therefore space space 2 space sin to the power of negative 1 end exponent space square root of fraction numerator 1 minus straight x over denominator 2 end fraction end root equals straight theta Again space space space 2 space cos to the power of negative 1 end exponent space square root of fraction numerator 1 plus straight x over denominator 2 end fraction end root equals 2 space cos to the power of negative 1 end exponent square root of fraction numerator 1 plus cos space straight theta over denominator 2 end fraction end root equals 2 space cos to the power of negative 1 end exponent square root of fraction numerator open parentheses 2 space cos squared begin display style straight theta over 2 end style close parentheses over denominator 2 end fraction end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 space cos to the power of negative 1 end exponent open parentheses cos straight theta over 2 close parentheses equals 2 open parentheses straight theta over 2 close parentheses equals straight theta therefore space space space space space space space 2 space cos to the power of negative 1 end exponent square root of fraction numerator 1 plus straight x over denominator 2 end fraction end root equals straight theta
From (1), (2) and (3), we get
    cos to the power of negative 1 end exponent space x space equals space 2 space sin to the power of negative 1 end exponent space square root of fraction numerator 1 minus x over denominator 2 end fraction end root equals 2 space cos to the power of negative 1 end exponent space square root of fraction numerator 1 plus x over denominator 2 end fraction end root

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation