Additional Mathematics 0606 / Trigonometry / 3.793.79: Derivative of reciprocal cosine0606/21/O/N/22 — Question 5 · 6 marksMark as done · Save for later · Show all solutions(a)You are given that y=1cos 2xy = \dfrac{1}{\mathrm{cos}\,2x}y=cos2x1. Show that dydx=k sin 2xcos22x\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{k\,\mathrm{sin}\,2x}{\mathrm{cos}^{2} 2x}dxdy=cos22xksin2x where kkk is a constant to be found.[2]▸ Answer(b)Find the values of xxx such that dydx=5sin 2x\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{5}{\mathrm{sin}\,2x}dxdy=sin2x5 for 0<x<π20 < x < \dfrac{\pi}{2}0<x<2π.[4]▸ Answer▸ Official mark scheme← 3.78: Exponential and logarithmic equations3.80: Counting 4-digit codes →
