Interim Assessment - 4 1β (a) Prove that β 2 (π ππθ+cosπππ θ ) −1 ≡2 tan π‘ππ θ. θ β β β β β [3] ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β (b) Hence solve the equation β 2 (π ππθ+cosπππ θ ) −1 θ = 5β° .β[3] ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… 1 [Turn Over] β 2 (a) Show that the equation 7tanπ‘ππ θ cosπππ θ + 12 = 0 can be expressed as 12θ − 7 sin π ππ θ − 12 = 0 .β β β β β [3] ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… (b)β Hence solve the equation 7tanπ‘ππ θ cosπππ θ + 12 = 0β° .β β β [3] ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β 2β 3β (a) Show that the equation cos πππ θ (7 tan π‘ππ θ − 5 cos πππ θ) = 1 can be written in the form πθ + π sin π ππ θ + π = 0, β β where a, b and c are integers to be found. [3] ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… (c)β Hence solve the equation cos πππ 2π₯ (7 tan π‘ππ 2π₯ − 5 cos πππ 2π₯) = 1 β° .β[3] ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β β 3 [Turn Over] 4β Find the exact solution of the equation 1 Π + (4π₯) = 6 1 − ( 2 3). β β β β β β β [2] ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β 4β ( 2 ) 5β (a) Verify the identity (2π₯ − 1) 4π₯ + 2π₯ − 1 β β 3 ≡ 8π₯ − 4π₯ + 1.β β [2] ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… (b) Prove the identity θ+1 θ−1 1 ≡ 1−2θ . β β β β β β β β β [3] ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β β 5 [Turn Over] c) Using the results of (a) and (b), solve the equation β β β β θ+1 θ−1 = 4 cos πππ θ β° .β β β [5] ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… β 6β
