Further Pure Mathematics 4PM1 / Trigonometry / 1.251.25: Past-paper question 114PM1/2R/June/2023 — Question 11 · 12 marksMark as done · Save for later(a) Use a formula on page 2 to show that sin2A=12(1−cos2A)\displaystyle \mathrm{sin}^2 A = \frac{1}{2}(1 - \mathrm{cos} 2A)sin2A=21(1−cos2A)(3)(b) Show that sin4x+cos4x=3+cos4x4\displaystyle \mathrm{sin}^4 x + \mathrm{cos}^4 x = \frac{3 + \mathrm{cos} 4x}{4}sin4x+cos4x=43+cos4x(5)(c) Hence solve, in degrees to one decimal place, the equation8sin4(θ2)+8cos4(θ2)=5sin(2θ)+6for0∘≤θ<180∘8\mathrm{sin}^4\left(\frac{\theta}{2}\right) + 8\mathrm{cos}^4\left(\frac{\theta}{2}\right) = 5\mathrm{sin}(2\theta) + 6 \quad \mathrm{for} 0^\circ \leq \theta < 180^\circ8sin4(2θ)+8cos4(2θ)=5sin(2θ)+6for0∘≤θ<180∘(4)▸ Mark scheme← 1.24: 4PM1/2R/June/2023 — Question 71.1: 4PM1/1/June/2022 — Question 3 →
