Target Mathematics

1.23: Trigonometric identities, trigonometric equations and exact trigonometric values

4PM1/2/June/2024 — Question 11 · 16 marks

Using formulae from page 2, show that
(a)(i)cos2A=2cos2A−1(a) (i) \mathrm{cos} 2A = 2\mathrm{cos}^2 A - 1
(3)
(ii)sin2A=2sinAcosA(ii) \mathrm{sin} 2A = 2\mathrm{sin} A \mathrm{cos} A
(1)
(b) Show that cos3A=cos3A+3cosA4(b) \text{ Show that } \mathrm{cos}^3 A = \frac{\mathrm{cos} 3A + 3\mathrm{cos} A}{4}
(4)
Hence, or otherwise,
(c) solve, giving exact values in terms of π\pi
8cos3(θ2)−6cos(θ2)−1=0for0≤θ≤2π8\mathrm{cos}^3\left(\frac{\theta}{2}\right) - 6\mathrm{cos}\left(\frac{\theta}{2}\right) - 1 = 0 \quad \mathrm{for} 0 \leq \theta \leq 2\pi
(4)
(d) use algebraic integration to find the exact value of
∫0π/6(4cos3θ−sin2θ) dθ\int_0^{\pi/6} (4\mathrm{cos}^3 \theta - \mathrm{sin} 2\theta) \, \mathrm{d}\theta
(4)