Zadanie 12c, Paper 1, listopad 2023, TZ1, Matematyka AA IB HL

(i) For those angles,5θ is π/2 or 3π/2, so cos5θ=0. Factor the identity as given and note both cosθ are nonzero (indeed positive); thus the quartic factor vanishes. (ii) Set u=cos²θ. Solving 16u²−20u+5=0 gives u=(5±√5)/8. Since cos decreases on (0,π/2), the larger value corresponds to π/10. The positive ratio R therefore satisfies $R^2=(5+\sqrt5)/(5-\sqrt5)=(3+\sqrt5)/2$. As $[(1+\sqrt5)/2]^2=(3+\sqrt5)/2$ and R>0, the asserted exact ratio follows.

(i) M1:use the factored identity. R1:both specified angles make cos5θ zero. R1:their cosines are nonzero, so the quartic factor must vanish; this conclusion depends on the preceding argument. (ii) M1:solve the quadratic in cos²θ. A1:both squared-cosine roots. M1:form the ratio with the larger root in the numerator and simplify its square. A1:verify the positive exact value (1+√5)/2. A valid alternate exact-root argument earns equivalent marks. The first R1 in (i), showing cos5θ=0 at the two angles, is independent of the M1; the final R1 depends on both. Alternative (ii): M1 treat the equation as a quadratic in squared cosines; A1 identify u+v=5/4, uv=5/16; M1 derive for R=√(u/v)>1 that R+1/R=(u+v)/√(uv)=√5; A1 solve for R=(√5+1)/2, choosing R>1. Credit relevant proof work already shown in another part without requiring repetition.