Using a Taylor series, find a quadratic approximation for \(f(x) = \sin x\) centred about \(x = \frac{{3\pi }}{4}\).

When using this approximation to find angles between \(130^\circ\) and \(140^\circ\), find the maximum value of the Lagrange form of the error term.

Hence find the largest number of decimal places to which \(\sin x\) can be estimated for angles between \(130^\circ\) and \(140^\circ\).

Explain briefly why the same maximum value of error term occurs for \(g(x) = \cos x\) centred around \(\frac{\pi }{4}\) when finding approximations for angles between \(40^\circ\) and \(50^\circ\).

\(f(x) = \sin x,{\text{ }}f'(x) = \cos x,{\text{ }}{f^{(2)}}(x) =  - \sin x\)     M1

\(f\left( {\frac{{3\pi }}{4}} \right) = \frac{1}{{\sqrt 2 }},{\text{ }}f'\left( {\frac{{3\pi }}{4}} \right) =  - \frac{1}{{\sqrt 2 }},{\text{ }}{f^{(2)}}\left( {\frac{{3\pi }}{4}} \right) =  - \frac{1}{{\sqrt 2 }}\)     A1

hence the quadratic Taylor Polynomial is

\(\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}\left( {x - \frac{{3\pi }}{4}} \right) - \frac{1}{{\sqrt 2 }}\frac{{{{\left( {x - \frac{{3\pi }}{4}} \right)}^2}}}{{2!}}\)     M1A1

\(\left( {\frac{1}{{\sqrt 2 }}\left( {1 - \left( {x - \frac{{3\pi }}{4}} \right) - \frac{1}{2}{{\left( {x - \frac{{3\pi }}{4}} \right)}^2}} \right)} \right)\)

\(f(x) = \sin x,{\text{ }}{f^{(3)}}(x) =  - \cos x\)     (A1)

the Lagrange form of the error term is: \(\left| {{R_n}(x)} \right| \leqslant \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{(n + 1)!}}\max \left| {{f^{n + 1}}(k)} \right|\)

\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x - \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\max \left| {{f^3}(k)} \right|\)     (M1)

\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x - \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\max \left| { - \cos k} \right|\)     A1

in this case \(\left| { - \cos k} \right| \leqslant \left| { - \cos 140} \right|\)     (A1)

\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x - \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\left| { - \cos 140} \right|\)

choosing \(140^\circ  = \frac{{14\pi }}{{18}}\)     M1

\( \Rightarrow \left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {\frac{{14\pi }}{{18}} - \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\left| { - \cos \frac{{14\pi }}{{18}}} \right|\)     A1

therefore the maximum value of the error term is \(8.48 \times {10^{ - 5}}\)     A1

\(\left| {{R_2}(x)} \right| \leqslant 8.48 \times {10^{ - 5}} = 0.0000848\) hence for angles between \(130^\circ\) and \(140^\circ\) the approximation will be accurate to 3 decimal places     A1

\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x - \frac{\pi }{4}} \right|}^3}}}{{3!}}\max \left| {\sin k} \right|\)     (M1)

since the max value of \(\left| {{f^3}(k)} \right|\) is \(\sin 50^\circ \) which is the same as \(\left| {\cos 140^\circ } \right|\)     A1R1

then the error is the same     AG

Part a) was answered successfully by most candidates. However, the majority of candidates struggled to gain full marks on the remainder of the question.

In part b) candidates struggled to work out which angle to use to find the maximum value.

Part a) was answered successfully by most candidates. However, the majority of candidates struggled to gain full marks on the remainder of the question.

In part d) most candidates understood that this was related to a translation of the sine graph but were unable to explain it convincingly.