Additional Mathematics 0606 / Trigonometry / 3.83.8: Normal to sine curve and triangle OAB0606/22/F/M/23 — Question 11 · 9 marksMark as done · Save for later · Show all solutionsThe normal to the curve y=sin(4x−π)y = \mathrm{sin}(4x - \pi)y=sin(4x−π) at the point A(a,0)A(a, 0)A(a,0), where π2<a<π\dfrac{\pi}{2} < a < \pi2π<a<π, meets the yyy-axis at the point BBB. Find the exact area of triangle OABOABOAB, where OOO is the origin.[9]▸ Answer▸ Official mark scheme← 3.7: Cosec–sec–tan simplifies to sine3.71: Modulus graphs and inequality →