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Using power series for $\sin{x}$ why do for some values of $x$ it needs to be worked out more than others? $$\sin{x}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^7}{7!}+...$$ From this lets say for the value of $\sin 1$ radian , it takes only 2 terms to get an approximate enough answer of $0.8333..$ But for $\sin 5$ radians it takes 7 terms to get an approximate answer. Why is it so? Till which terms would be plausible to use in my exams if I really needed to calculate sin values(they don't allow calculators)

J. W. Tanner
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Naveen V
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    Related: 1338411 and 105593 – user170231 Feb 27 '23 at 18:46
  • I will look into it , thank you ! – Naveen V Feb 27 '23 at 18:48
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    No calculator-banning exam will ask you for $\sin5$. – J.G. Feb 27 '23 at 18:49
  • Intuitively, the convergent sequence will not start to converge until the terms in the sequence are very close to $0$. Consider the sequence of terms $$\frac{5^1}{1!}, ~\frac{5^3}{3!}, ~\frac{5^5}{5!}, ~\frac{5^7}{7!}, ~\frac{5^9}{9!}.$$ The first few terms are not very close to $0$. – user2661923 Feb 27 '23 at 19:03
  • ah that makes sense intuitively @user2661923 for me so for values close to 0 the first few terms are plausible and higher the value higher number of terms are needed to get an approx – Naveen V Feb 27 '23 at 19:07
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    As others noted, smaller argument will lead to quicker convergence. So use redaction formulas! $\sin 5=\sin(5-2\pi) \approx -\sin 1.28$ – Vasili Feb 27 '23 at 19:23
  • Thats a good idea too ! @Vasili thank you – Naveen V Feb 27 '23 at 19:27

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