How do we show 1−cos⁡x≥x23 for |x|≤1? My first idea was to write1−cos⁡x=12|eix−1|2which is true for all x∈R, but I don't have a suitable lower bound for the right-hand side at hand.

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How do we show 1−cos⁡x≥x23 for |x|≤1? My first idea was to write1−cos⁡x=12|eix−1|2which is true for all x∈R, but I don&#039;t have a suitable lower bound for the right-hand side at hand.

Ethan Sanders · Accepted Answer

For |x|≤1 the Taylor series  cos⁡(x)=1−x22!+x44!−x66!+…  is an alternating series with terms that decrease in absolute value. It follows that for these x  cos⁡(x)≤1−x22!+x44!≤1−x22!+x24!=1−1124x2  and therefore  1−cos⁡(x)≥1124x2  That is an even better estimate since 1124&amp;gt;13  The same approach can be used to show that 1−cos⁡(x)≥x23 holds on the larger interval [−2,2]:  cos⁡(x)≤1−x22!+x44!≤1−x22!+4x24!=1−13x2  because the x2n(2n)! terms decrease in absolute value for n≥1

Cheryl King · Accepted Answer

Since sinx is concave for x∈[0,π2], if x∈[0,1] then sin⁡x≤xsin⁡1⇒1−cos⁡x≥x212sin⁡1. The x2 coefficient approximates 0.42. The generalization x∈[−1,1] follows from 1−cos⁡x,cx2 being even.

Vasquez · Accepted Answer

f(x)=1−cos⁡xx2 is an even function, so we only look at x∈[0,1]. It&#039;s easy to prove f(x) is continuous and differentiable at x=0, and f′(x)=xsin⁡x+2cos⁡x−2x3 Now xsin⁡x+2cos⁡x−2≤2tan⁡x2⋅sin⁡x−4sin2⁡x2 =2tan⁡x2⋅2sin⁡x2cos⁡x2−4sin2⁡x2=0,∀x∈[0,1] Hence f(x)≥f(1)=1−cos⁡1≈.46&amp;gt;13,  1−cos⁡x≥(1−cos⁡1)x2&amp;gt;13x2

How do we show 1-\cos x \geq \frac{x^2}{3} for |x|

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