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

  • MHB
  • Thread starter Amer
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In summary, the conversation discusses the relationship between x = sin x at x = 0 and the slope of each curve, 1 and cos x. It is mentioned that sin(x) will always decrease and x will remain in its direction, and the two curves will never meet unless the slope of sin(x) is greater than 1. The speaker also brings up the possibility of using the Intermediate Value Theorem to further explore this relationship.
  • #1
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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  • #2
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. [tex]-1 \le \sin(x) \le 1 \therefore[/tex], For [tex]x>1[/tex], we have [tex]x > \sin(x)[/tex].

Now we can focus on [tex]0 < x \le 1[/tex].

2) I might be tempted to to ponder [tex]f(x) = x - \sin(x)[/tex]. There must be an Intermediate Value Theorem, or something, in there.
 
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FAQ: Does sinx < x hold true for all positive values of x?

What is the meaning of "Prove that sinx < x for x>0"?

The statement means to mathematically show that for any positive value of x, the value of sine of x is always less than x.

Why is it important to prove that sinx < x for x>0?

This inequality is a fundamental concept in trigonometry and calculus, and it is used in various mathematical proofs and applications. Understanding this inequality can also help in understanding the behavior of trigonometric functions.

How can you prove that sinx < x for x>0?

The proof involves using the Taylor series expansion of sine function and analyzing the remainder term to show that it is always positive for x>0. This implies that the value of sine function is always less than the value of its argument, x.

Can you provide an example to illustrate this inequality?

For example, if we take x=1, then sin1=0.8414 and 1>0.8414. Similarly, for any positive value of x, the value of sine function will always be less than x.

Does this inequality hold true for negative values of x?

No, the inequality does not hold true for negative values of x. In fact, for negative values, the opposite inequality holds, i.e. sinx > x. This is because the sine function is an odd function, which means it behaves differently for positive and negative values of x.

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