Target Mathematics

1.30: Trigonometric identities

4PM1/1/June/2023 — Question 9 · 13 marks

(a) Using the formulae on page 2, show that
(i)cos2A=cos2A+12(i) \mathrm{cos}^2 A = \frac{\mathrm{cos} 2A + 1}{2}
(ii)sin2A=1−cos2A2(ii) \mathrm{sin}^2 A = \frac{1 - \mathrm{cos} 2A}{2}
(4)
(b) Show that
(2sinx−cosx)(sinx−3cosx)=12(cos2x−7sin2x+5)(2 \mathrm{sin} x - \mathrm{cos} x)(\mathrm{sin} x - 3 \mathrm{cos} x) = \frac{1}{2} (\mathrm{cos} 2x - 7 \mathrm{sin} 2x + 5)
(5)
y=(2sinx−cosx)(sinx−3cosx)y = (2 \mathrm{sin} x - \mathrm{cos} x)(\mathrm{sin} x - 3 \mathrm{cos} x)
(c) Solve, for 0∘≤x≤180∘0^\circ \leq x \leq 180^\circ the equation, dydx=0\frac{\mathrm{d}y}{\mathrm{d}x} = 0
Give your answers to the nearest whole number.
(4)