Additional Mathematics 0606 / Trigonometry / 3.153.15: Sin³ identity and triple-angle equation0606/22/F/M/25 — Question 8 · 7 marksMark as done · Save for later · Show all solutions(a)Show that sin θ tan2θsec2θ\dfrac{\mathrm{sin}\,\theta\,\mathrm{tan}^{2}\theta}{\mathrm{sec}^{2}\theta}sec2θsinθtan2θ can be written as sin3θ\mathrm{sin}^{3}\thetasin3θ.[2]▸ Answer(b)Hence solve sin 3x tan23xsec23x=18\dfrac{\mathrm{sin}\,3x\,\mathrm{tan}^{2} 3x}{\mathrm{sec}^{2} 3x} = \dfrac{1}{8}sec23xsin3xtan23x=81 for −180∘⩽x⩽180∘-180^{\circ} \leqslant x \leqslant 180^{\circ}−180∘⩽x⩽180∘.[5]▸ Answer▸ Official mark scheme← 3.14: Composite exponential function3.95: Cosine graph parameters from diagram →
