Prove that(sin2x2)⋅433⋅1sin⁡x+1−x23}&lt;1

Note that for x∈[π2,π], then 12≤sin2x2≤1Now consider the function f(x)=sin⁡x+1−xFor x=π2 then f(x)&gt;0 and for x=π then f(x)&lt;0 so, since f is continuous, there is a zero of f in (π2,π). Let a be that zero.Now let g(x)=1f(x)23 Note that g(x)&gt;0 since g(x)=(f(x)−13)2. If x is in a neighbourhood of a then g(x) has no upper bound. Therefore:(sin2x2)⋅433⋅1sin⁡x+1−x23}≥12⋅433⋅g(x)But g(x) has no upper bound so in a small enough neighbourhood of (sin2x2)⋅433⋅1sin⁡x+1−x23}&gt;1

"Prove that(sin2x2)⋅433⋅1sin⁡x+1−x23}<1" was answered by our community

michiiiiiakqm

Answered question

2022-03-31

Prove that

(\sin^{2} \frac{x}{2}) \cdot \frac{\sqrt[3]{4}}{3} \cdot {\frac{1}{\sin x + 1 - x}}^{\frac{2}{3}}} < 1

Answer & Explanation

armejantm925

Beginner2022-04-01Added 20 answers

Note that for

x \in [\frac{π}{2}, π]

, then

\frac{1}{2} \leq \sin^{2} \frac{x}{2} \leq 1

Now consider the function

f (x) = \sin x + 1 - x

For

x = \frac{π}{2}

then

f (x) > 0

and for

x = π

then

f (x) < 0

so, since f is continuous, there is a zero of f in

(\frac{π}{2}, π)

. Let a be that zero.
Now let

g (x) = \frac{1}{{f (x)}^{\frac{2}{3}}}

Note that

g (x) > 0

since

g (x) = {({f (x)}^{- \frac{1}{3}})}^{2}

. If x is in a neighbourhood of a then g(x) has no upper bound. Therefore:

(\sin^{2} \frac{x}{2}) \cdot \frac{\sqrt[3]{4}}{3} \cdot {\frac{1}{\sin x + 1 - x}}^{\frac{2}{3}}} \geq \frac{1}{2} \cdot \frac{\sqrt[3]{4}}{3} \cdot g (x)

But g(x) has no upper bound so in a small enough neighbourhood of

(\sin^{2} \frac{x}{2}) \cdot \frac{\sqrt[3]{4}}{3} \cdot {\frac{1}{\sin x + 1 - x}}^{\frac{2}{3}}} > 1

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