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Inverse Trigonometric Functions

Question
CBSEENMA12032648

Find the principal values of the following

cos to the power of negative 1 end exponent open parentheses negative fraction numerator 1 over denominator square root of 2 end fraction close parentheses


Solution
Let space space space straight y space equals cos to the power of negative 1 end exponent open parentheses negative fraction numerator 1 over denominator square root of 2 end fraction close parentheses space space where space 0 less or equal than straight y less or equal than straight pi
therefore space space space space space cos space straight y space equals space minus fraction numerator 1 over denominator square root of 2 end fraction space space where space 0 less or equal than straight y less or equal than straight pi
therefore space space space space space cos space straight y space equals negative cos straight pi over 4 equals space cos space open parentheses straight pi minus straight pi over 4 close parentheses equals space cos space fraction numerator 3 straight pi over denominator 4 end fraction
therefore space space space space space straight y equals fraction numerator 3 straight pi over denominator 4 end fraction
therefore space space space space space  required principal value = fraction numerator 3 straight pi over denominator 4 end fraction

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