C3 Exam Questions based on Chapter 6 only (not Chapter 7) [June 2005 Q1] 1. (a) Given that sin2 + cos2 1, show that 1 + tan2 sec2 . (2) (b) Solve, for 0 < 360, the equation 2 tan2 + sec = 1, giving your answers to 1 decimal (6) [June 2013 (Withdrawn) Q2] 2. (note part (c) requires Chp 7 knowledge) [Jan 2012 Q5] 3. [Jan 2010 Q8] 4. Solve cosec2 2x cot 2x = 1 – for 0 180. x (7) (a) Use the identity cos2 + sin2 = 1 to prove that tan2 = sec2 1. [June 2009 Q2] 5. θ θ θ θ – (2) (b) for 0 < 360°, the equation θ 2 tan2 + 4 sec + sec2 = 2. θ θ θ (6) [June 2008 Q5] 6. (a) Given that sin2 + cos2 1, show that 1 + cot2 cosec2 . θ θ ≡ θ ≡ θ (2) (b) Solve, for 0 < 180°, the equation θ 2 cot2 9 cosec = 3, θ – θ giving your answers to 1 decimal place. (6) [Jan 2008 Q6b] 7. (i) Prove that + 2 sec (2n + 1). x, x ≠ (4) (ii) Hence find, for 0 < < 2π, all the solutions of x + = 4. (3)
C3 Exam Questions based on Chapter 6 only (not Chapter 7) [June 2005 Q1] 1. (a) Given that sin2 + cos2 1, show that 1 + tan2 sec2 . (2) (b) Solve, for 0 < 360, the equation 2 tan2 + sec = 1, giving your answers to 1 decimal (6) [June 2013 (Withdrawn) Q2] 2. (note part (c) requires Chp 7 knowledge) [Jan 2012 Q5] 3. [Jan 2010 Q8] 4. Solve cosec2 2x – cot 2x = 1 for 0 x 180. (7) [June 2009 Q2] 5. (a) Use the identity cos2 θ + sin2 θ = 1 to prove that tan2 θ = sec2 θ – 1. (2) (b) for 0 θ < 360°, the equation 2 tan2 θ + 4 sec θ + sec2 θ = 2. (6) [June 2008 Q5] 6. (a) Given that sin2 θ + cos2 θ ≡ 1, show that 1 + cot2 θ ≡ cosec2 θ . (2) (b) Solve, for 0 θ < 180°, the equation 2 cot2 θ – 9 cosec θ = 3, giving your answers to 1 decimal place. (6) [Jan 2008 Q6b] 7. (i) Prove that + 2 sec x, x ≠ (2n + 1). (4) (ii) Hence find, for 0 < x < 2π, all the solutions of + = 4. (3)
C3-Chp6-TrigExamQuestions.docx