A-LevelMathematicsTrigonometry9709/21 · Oct/Nov 2023 · Q69 Marks
Show that cosec θ(3 sin 2θ + 4 sin³ θ) = 4 + 6 cos θ – 4 cos² θ.
- (a)Show that cosec θ(3 sin 2θ + 4 sin³ θ) = 4 + 6 cos θ – 4 cos² θ.[3]
- (b)Solve the equation cosec θ(3 sin 2θ + 4 sin³ θ) + 3 = 0 for –π < θ < 0.[3]
- (c)Find ∫ cosec θ(3 sin 2θ + 4 sin³ θ) dθ.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.
- [B1] Use cosecθ = 1/sinθ
- [M1] Express in terms of sine and cosine only Dependent on B1.
- [A1] Obtain given result 4+6cosθ-4cos²θ with sufficient detail AG
- [M1] Attempt use of formula to solve 3-term quadratic equation as far as cosθ = k₁ where -1<k₁ <1.
- [A1] Solve 4cos²θ−6cosθ-7=0 to obtain at least cosθ=−0.770... or exact equivalent cosθ = (6-√148)/8
- [A1] Obtain −2.45 or greater accuracy; and no others between –π and 0.
- [M1] Express cos²θ term in the form k₂ + k₃ cos 2θ where k₂k₃ ≠ 0.
- [A1] Obtain integrand 6cosθ+2-2cos2θ Following the 3-term expression in cosθ from part (a).
- [A1] Integrate to obtain correct 6sinθ+2θ-sin2θ Condone absence of + c, but all in terms of θ.
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