Inverse Trigonometric Functions

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Question

Shoe that $sin to the power of negative 1 end exponent open parentheses 2 space x square root of 1 minus x squared end root close parentheses equals 2 space cos to the power of negative 1 end exponent space x comma space fraction numerator 1 over denominator square root of 2 end fraction space less or equal than x less or equal than 1$

Solution

Pur space straight x space equals space straight theta space therefore space space straight L. straight H. straight S. space equals space sin to the power of negative 1 end exponent open parentheses 2 space straight x space square root of 1 minus straight x squared end root close parentheses equals sin to the power of negative 1 end exponent open parentheses 2 space cos space straight theta space square root of 1 minus cos squared space straight theta end root close parentheses space space space space space space space space space space space space space space space space equals space sin to the power of negative 1 end exponent left parenthesis 2 space cos space straight theta space sin space straight theta right parenthesis space equals space sin to the power of negative 1 end exponent space left parenthesis sin space 2 space straight theta right parenthesis space space space space space space space space space space space space space space space space equals space 2 space straight theta space equals space 2 space cos to the power of negative 1 end exponent space straight x space equals space straight R. straight H. straight S. therefore space space space space straight L. straight H. straight S. space equals space straight R. straight H. straight S.