MHB Does sinx < x hold true for all positive values of x?

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The discussion centers on whether sin(x) is less than x for all positive values of x. It is established that at x = 0, sin(x) equals x, but as x increases, the slope of sin(x) (cos(x)) is always less than or equal to 1, indicating that sin(x) will not catch up to x for x > 0. The argument suggests limiting the proof to the range 0 < x ≤ 1, where it is noted that sin(x) remains bounded between -1 and 1. The concept of analyzing the function f(x) = x - sin(x) is introduced, hinting at the potential application of the Intermediate Value Theorem. Overall, the conclusion leans towards the assertion that sin(x) is indeed less than x for all positive x.
Amer
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My work it is clear that
x = sin x, at x = 0
if we look at the slope of each curve

1 , cos x
cos x <= 1
which means sin(x) will go down and x will remain at it is direction, and the two curves will never met unless the slope of sin(x) be more than 1
How about it ?
Thanks in advance
 
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It is never good to start with "It is clear that...". Really, just never do that. Simply state what you wish to state. No need to insult your audience.

Your argument is fuzzy, but you are on the right track.

1) I am tempted to limit the scope of the proof by pointing out one thing before I start the actual proof. -1 \le \sin(x) \le 1 \therefore, For x&gt;1, we have x &gt; \sin(x).

Now we can focus on 0 &lt; x \le 1.

2) I might be tempted to to ponder f(x) = x - \sin(x). There must be an Intermediate Value Theorem, or something, in there.
 
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