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

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Solution

Put x = sin θ

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

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