Prove that x−x33!&lt;sin⁡(x)&lt;x for all x&gt;0

Recursive integration We know that0≤cos⁡a≤1⇒sin⁡t=∫0tcos⁡sds&lt;tfor 0&lt;t&lt;z&lt;x. Integrating over (0,z) we get1−cos⁡z=∫0zsin⁡tdt&lt;∫0ztdt=z22that is for all 0&lt;z&lt;x we have,1−z22&lt;cos⁡z≤1integrating again over (0,x) we getx−x36=∫0x1−z22dz&lt;∫0xcos⁡zdz=sin⁡xthat is x−x36&lt;sin⁡x&lt;xcontinuing with this process you get,1−x22&lt;cos⁡x&lt;1−x22+x424x−x36&lt;sin⁡x&lt;x−x36+x55!More generally for n≥1, by induction we get∑k=02n−1(−1)kx2k(2k)!&lt;cos⁡x&lt;∑k=02n−1(−1)kx2k(2k)!+x4n(4n)!∑k=02n−1(−1)kx2k+1(2k+1)!&lt;sin⁡x&lt;∑

Expert Answer to "Prove that x−x33!<sin⁡(x)<x for all x>0"

klamytewoc

Answered question

2022-03-01

Prove that

x - \frac{x^{3}}{3!} < \sin (x) < x

for all x>0

Answer & Explanation

OxicegeWheegeb0k

Beginner2022-03-02Added 3 answers

Recursive integration We know that

0 \leq \cos a \leq 1 \Rightarrow \sin t = \int_{0}^{t} \cos s d s < t

for

0 < t < z < x

. Integrating over

(0, z)

we get

1 - \cos z = \int_{0}^{z} \sin t d t < \int_{0}^{z} t d t = \frac{z^{2}}{2}

that is for all

0 < z < x

we have,

1 - \frac{z^{2}}{2} < \cos z \leq 1

integrating again over

(0, x)

we get

x - \frac{x^{3}}{6} = \int_{0}^{x} 1 - \frac{z^{2}}{2} d z < \int_{0}^{x} \cos z d z = \sin x

that is

x - \frac{x^{3}}{6} < \sin x < x

continuing with this process you get,

1 - \frac{x^{2}}{2} < \cos x < 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24}

x - \frac{x^{3}}{6} < \sin x < x - \frac{x^{3}}{6} + \frac{x^{5}}{5!}

More generally for

n \geq 1

, by induction we get

\sum_{k = 0}^{2 n - 1} {(- 1)}^{k} \frac{x^{2 k}}{(2 k)!} < \cos x < \sum_{k = 0}^{2 n - 1} {(- 1)}^{k} \frac{x^{2 k}}{(2 k)!} + \frac{x^{4 n}}{(4 n)!}

\sum_{k = 0}^{2 n - 1} {(- 1)}^{k} \frac{x^{2 k + 1}}{(2 k + 1)!} < \sin x < \sum

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