Find the value of γ€–sinγ€—^(-1) [cos(33π/5)]

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

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Transcript

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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo