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

  1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Serial order wise

Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.