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

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SUMMARY

The discussion confirms that the inequality sin(x) < x holds true for all positive values of x. The proof begins with the observation that at x = 0, both functions intersect, but as x increases, the slope of sin(x), represented by cos(x), remains less than or equal to 1, ensuring that sin(x) will not exceed x for x > 0. The argument is further solidified by noting that for x > 1, sin(x) is bounded between -1 and 1, reinforcing that x will always be greater than sin(x) in this range. The focus on the interval 0 < x ≤ 1 allows for a more detailed examination of the behavior of the functions.

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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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