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

Misc 12 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Misc 12 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3
Misc 12 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4

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Misc 12 Prove 9π/8 – 9/4 sin−1 1/3 = 9/4 sin−1 (2√2)/3 Solving 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 Using sin-1x + cos−1x = 𝝅/𝟐 cos-1x = 𝜋/2 – sin−1x Replace x by 1/3 cos-1 1/3 = 𝜋/2 – sin−1 1/3 We convert cos−1 to sin−1 Let a = "cos−1" 1/3 cos a = 1/3 Now, sin a = √(1−cos2 𝑎) =√(1−(1/3)^2 ) "=" √(1−1/9) "=" √((9 − 1)/9) "=" √(8/9)=√((22 × 2)/32) "=" (√(2^2 ) × √2)/√(3^2 ) "=" (2 √2)/3 Thus, 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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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