excluderho

2022-06-27

Please could somebody explain how the expression involving $\theta $ that

$\frac{1+\mathrm{sin}\theta}{5+3\mathrm{tan}\theta -4\mathrm{cos}\theta}$

approximates to for small values of $\theta $ is $1-2\theta +4{\theta}^{2}$?

$\frac{1+\mathrm{sin}\theta}{5+3\mathrm{tan}\theta -4\mathrm{cos}\theta}$

approximates to for small values of $\theta $ is $1-2\theta +4{\theta}^{2}$?

Cahokiavv

Beginner2022-06-28Added 31 answers

Just compute the first terms of the Taylor series of $\frac{1+\mathrm{sin}\theta}{5+3\mathrm{tan}\theta -4\mathrm{cos}\theta}$, and you will get $1-2\theta +4{\theta}^{2}$.

misurrosne

Beginner2022-06-29Added 3 answers

Approximate the numerator and denominator up to ${\theta}^{2}$ terms and get

$\frac{1+\theta +o({\theta}^{2})}{1+3\theta +2{\theta}^{2}+o({\theta}^{2})}=(1+\theta )(1-3\theta +7{\theta}^{2})+o({\theta}^{2})=1-2\theta +4{\theta}^{2}+o({\theta}^{2}).$

$\frac{1+\theta +o({\theta}^{2})}{1+3\theta +2{\theta}^{2}+o({\theta}^{2})}=(1+\theta )(1-3\theta +7{\theta}^{2})+o({\theta}^{2})=1-2\theta +4{\theta}^{2}+o({\theta}^{2}).$

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