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Misc 12 - Chapter 2 Inverse Trigonometry - Prove 9pi/8 - 9/4 - Miscellaneous

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Misc 12 Prove 9π/8 – 9/4 sin-1 1/3 = 9/4 sin-1 (2√2)/3 Taking L.H.S. 9π/8 – 9/4 sin-1 1/3 = 9/4 (π/(2 )−"sin−1 " 1/3) = 9/4 "cos−1 " 1/3 Let a = "cos−1" 1/3 cos a = 1/3 Now, sin2 a = 1 – cos2 a sin a = √(1−cos2 𝑎) =√(1−(1/3)^2 ) "=" √(1−1/9) "=" √((9 − 1)/9) "=" √(8/9) "=" √((4 × 2)/32) " =" √((22 × 2)/32) "=" (√(2^2 ) × √2)/√(3^2 ) "=" (2 √2)/3 ∴ "sin a =" (2 √2)/3 a = sin-1 ((2 √2)/3) Hence, "cos−1 " 1/3 = a = sin-1 ((2 √2)/3) Now, From (1) 9π/8 – 9/4 sin-1 1/3 = 9/4 "cos−1 " 1/3 Putting value = 9/4 sin-1 ((2 √2)/3) Hence, 9π/8 – 9/4 sin-1 1/3 = 9/4 sin-1 ((2 √2)/3) Hence proved

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