Inverse Trigonometric Functions

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Using principle value, evaluate the following : - Wired Faculty

Question

Using principle value, evaluate the following :

$cos to the power of negative 1 end exponent open parentheses cos fraction numerator 2 straight pi over denominator 3 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses sin fraction numerator 2 straight pi over denominator 3 end fraction close parentheses.$

Solution

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