Find the principal values of the following from Mathematics Inv
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    Inverse Trigonometric Functions
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    Find the principal values of the following
    cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses
    Let equals cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses equals space straight y space where space 0 space less or equal than space straight y space less or equal than space straight pi therefore space space space space cos space straight y space equals space minus 1 half space space where space 0 space less or equal than space straight y space less or equal than space straight pi rightwards double arrow space space space space space straight y equals fraction numerator 2 straight pi over denominator 3 end fraction space space space space space space space space space space space space space space space space space space open square brackets because space space space cos fraction numerator 2 straight pi over denominator 3 end fraction equals cos space open parentheses straight pi minus straight pi over 3 close parentheses equals negative cos straight pi over 3 equals negative 1 half close square brackets

    therefore     required principal value = fraction numerator 2 straight pi over denominator 3 end fraction
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    Answer
    Find the principal values of the following
    cosec-1(2)

    Let y == cosec-1(2)               where straight y element of open square brackets fraction numerator negative straight pi over denominator 2 end fraction comma space 0 close square brackets union open parentheses 0 comma space straight pi over 2 close parentheses

    therefore   cosec y = 2                    where straight y element of open square brackets fraction numerator negative straight pi over denominator 2 end fraction comma space 0 close square brackets union open parentheses 0 comma space straight pi over 2 close parentheses
    therefore space space space space space space space space space space space space space space space space space space space y equals straight pi over 6

    therefore     required  prinicipal value = straight pi over 6

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    Find the principal values of the following
    sin to the power of negative 1 end exponent open parentheses negative 1 half close parentheses

    Let space straight y space equals space sin to the power of negative 1 end exponent open parentheses negative 1 half close parentheses space space where space minus straight pi over 2 space less or equal than space straight y space less or equal than space straight pi over 2 therefore space space sin space straight y space equals space minus 1 half space space space space space space space where space minus straight pi over 2 space less or equal than space straight y space less or equal than space straight pi over 2 therefore space space straight y equals negative straight pi over 6 space space space space space space space space space space space space space space open square brackets because space sin space open parentheses negative straight pi over 6 close parentheses equals negative sin straight pi over 6 equals negative 1 half close square brackets

    therefore required principal value = space minus straight pi over 6
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    Find the principal values of the following
    tan to the power of negative 1 end exponent left parenthesis negative square root of 3 right parenthesis
    Let space straight y space equals space tan to the power of negative 1 end exponent left parenthesis negative square root of 3 right parenthesis space space space space space space space space space where space minus straight pi over 2 less than straight y less than straight pi over 2 therefore space space space tan space y space equals space minus square root of 3 space space space space space space space space space space space space where space minus straight pi over 2 space less than straight y less than straight pi over 2 therefore space space space tan space straight y space equals space minus tan straight pi over 3 equals tan open parentheses negative straight pi over 3 close parentheses space where space minus straight pi over 2 less than straight y less than straight pi over 2 therefore space space space space straight y equals negative straight pi over 3
    therefore   required principal value = negative straight pi over 3.
    204 Views
    Answer
    Find the principal values of the following
    cos to the power of negative 1 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses
    Let space straight y equals cos to the power of negative 1 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses space space space space space space space where space 0 less or equal than straight y less or equal than straight pi therefore space space space cos space straight y equals fraction numerator square root of 3 over denominator 2 end fraction space space space space space space space space space space space where space 0 space less or equal than space straight y less or equal than straight pi therefore space space space space space straight y equals straight pi over 6
    therefore  required principal value = straight pi over 6
    204 Views
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