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Find the value of γ€–sinγ€—^(-1) [cos(33π/6)]

This question is similar to Example-9 Chapter-2 Inverse trigonometry



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sinβˆ’1 ("cos " 33Ο€/6) = sinβˆ’1 ("cos " (πŸ”π…+πŸ‘π…/πŸ“)) = sinβˆ’1 ("cos " 3πœ‹/5) = sinβˆ’1 ("sin" (𝝅/𝟐 βˆ’πŸ‘π›‘/πŸ“)) = sinβˆ’1 ("sin" ((5πœ‹ βˆ’ 6πœ‹)/10 )) = sinβˆ’1 ("sin" ((βˆ’π…)/𝟏𝟎 )) Let y = sinβˆ’1 ("sin" ((βˆ’πœ‹)/10 )) sin y = "sin" ((βˆ’πœ‹)/10 ) sin y = sin (-18Β°) Hence, y = (βˆ’π…)/𝟏𝟎 Which is in the range of sin-1 i.e. [(βˆ’π›‘)/𝟐, 𝛑/𝟐] Hence, γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) [𝒄𝒐𝒔(πŸ‘πŸ‘π…/πŸ”)] = (βˆ’π…)/𝟏𝟎 Hence, γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) [𝒄𝒐𝒔(πŸ‘πŸ‘π…/πŸ”)] = (βˆ’π…)/𝟏𝟎

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.