Question

3 (b) f(1) = esin(1-1) We have → g(x) = sin(1 - 0) g'(x) = – cos(1 – x) → (0= sin(1), s'(0) = – cos(1), SO sin(1 - 1) = sin(1) - cos(1)x+ O(x²). We can therefore write f(x) = sin(1)-cos(1)x+0(x2) = esin(l)e-cos(1)x+0(x²) = esin(1) [1 – cos(1)x+O(x²) = esin(1) – esin(1) cos(1)x+(x?). (4)

so this is the solution to the problem but im still confused on how they got from the second line to the third line and on. Please help, explain and show all work

Answer 1

We have the function f(x) - Sin Cix) also gcx) = sin(1-x) , so gco) = sin(4) and g'(x) = – cos (1-2) and g'co) = - COSCI) We expand sin (1-x) in Taylor's series about xso, then we get Sin (1-x) a Sin ( 1 0 + 1 d sin (1-x) a (no) + 0(2²) other terms containing x²] The approximation sign is due to that we consider only upto this in the series. Therefore, Sim (1-x) = sin(1) - Cos (1) x + (x²) Hence we may write, f(x) = e Sin (1-x) - Sin (1) - Cos (1) x +0 (x2) => f(1) = elan (1) e-Cos (1) x +o(a) >fca) = esined) [ e-Cer(1) 44 064)]. Now, ex = 4x + 2 y 2 to, using thill in the above, we get f(x= pelin (I) [ 1 - {cos (1)2 + 0(2}}}] (comanda ang uto) fece) = sin() [ ! - Cos (1) x + 0(e)] Piothere in the last term & we resolve the negative Sign within 0(x2) 1 There fore, f(x) = (fin (o) - elin (1) cos(L) x + (x2) This is the required answer,