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CBSE Class 12-science Answered

Evaluate sin left parenthesis 2 cos to the power of negative 1 end exponent left parenthesis negative 3 over 5 right parenthesis right parenthesis
Asked by Navya Benny | 10 Jan, 2016, 10:06: PM
answered-by-expert Expert Answer
L e t space cos to the power of negative 1 end exponent open parentheses negative 3 over 5 close parentheses equals theta rightwards double arrow cos theta equals negative 3 over 5 rightwards double arrow theta element of left parenthesis straight pi over 2 comma straight pi right square bracket Hence comma space sinθ space is space positive. sinθ equals square root of 1 minus open parentheses negative 3 over 5 close parentheses squared end root equals 4 over 5 Hence comma space sin 2 straight theta equals 2 sinθcosθ equals 2 cross times 4 over 5 cross times open parentheses negative 3 over 5 close parentheses equals negative 24 over 25 therefore sin open parentheses 2 cos to the power of negative 1 end exponent open parentheses negative 3 over 5 close parentheses close parentheses equals negative 24 over 25
Answered by satyajit samal | 11 Jan, 2016, 12:36: PM

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