 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/15/c/ii/solution

Put

h = x - π /2

. The sine addition identity gives

sin x = cos h = 1 - h^{2} /2 + h^{4} /24 + O (h^{6})

. Applying the exponential series gives

e^{s i n x} = e exp (- \frac{h ^{2}}{2} + \frac{h ^{4}}{24} + O (h^{6})) = e [1 - \frac{h ^{2}}{2} + (\frac{1}{24} + \frac{1}{8}) h^{4} + O (h^{6})] .

Thus the first three nonzero terms in the Taylor series about

π /2

are

e^{s i n x} = e - \frac{e}{2} (x - \frac{π}{2})^{2} + \frac{e}{6} (x - \frac{π}{2})^{4} + O ((x - \frac{π}{2})^{6}) .

 New to topics? Read the docs here!