Inverse Trigonometric Functions

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Find the value of sin –1 - Wired Faculty

Question

Find the value of sin^–1 $open parentheses sin space fraction numerator 3 straight x over denominator 5 end fraction close parentheses.$

Solution

sin to the power of negative 1 end exponent open parentheses sin space fraction numerator 3 straight pi over denominator 5 end fraction close parentheses equals sin to the power of negative 1 end exponent open square brackets sin open parentheses straight pi minus fraction numerator 2 straight pi over denominator 5 end fraction close parentheses close square brackets space space 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 open square brackets sin fraction numerator 2 straight pi over denominator 5 end fraction close square brackets equals fraction numerator 2 straight pi over denominator 5 end fraction

Note space colon space sin to the power of negative 1 end exponent left parenthesis sin space straight x right parenthesis equals straight x space where space straight x element of open square brackets negative straight pi over 2 comma space straight pi over 2 close square brackets space space space space space space space Now space sin to the power of negative 1 end exponent space open parentheses sin space fraction numerator 3 straight pi over denominator 5 end fraction close parentheses not equal to space fraction numerator 3 straight pi over denominator 5 end fraction as fraction numerator 3 straight pi over denominator 5 end fraction not an element of open square brackets negative straight pi over 2 comma space straight pi over 2 close square brackets 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