Exploring the Limits of Trigonometry: 0 < sin x < x

[Leong]

Given that 0 < sin x < x is true for 0 < x < π/2.
From the above, can we conclude that 0 < sin (x/2) < x/2? How about 0 < sin (x/5) < x/5? Why?
How about 0<sin 3x < 3x ? Why?

 

[mfb]

For suitable ranges of x, sure.

If you replace x by x/2 everywhere consistently, you don't change anything, you just replaced your variable. An analogy would be to replace all "x" by "y".

0 < sin y < y is true for 0 < y < π/2
Now define y=x/2.

 

[Leong]

For suitable ranges of x, sure.

If you replace x by x/2 everywhere consistently, you don't change anything, you just replaced your variable. An analogy would be to replace all "x" by "y".

0 < sin y < y is true for 0 < y < π/2
Now define y=x/2.

Thank you very much for the explanation.

 

[PeroK]

Given that 0 < sin x < x is true for 0 < x < π/2.
From the above, can we conclude that 0 < sin (x/2) < x/2? How about 0 < sin (x/5) < x/5? Why?
How about 0<sin 3x < 3x ? Why?

Yes, but you also have to change your range:

$0 < \sin x < x$ for $0 < x < \pi/2$

Is equivalent to:

$0 < \sin x/2 < x/2$ for $0 < x < \pi$

 

[Leong]

Yes, but you also have to change your range:

$0 < \sin x < x$ for $0 < x < \pi/2$

Is equivalent to:

$0 < \sin x/2 < x/2$ for $0 < x < \pi$

 

[WWGD]

Given that 0 < sin x < x is true for 0 < x < π/2.
From the above, can we conclude that 0 < sin (x/2) < x/2? How about 0 < sin (x/5) < x/5? Why?
How about 0<sin 3x < 3x ? Why?

Essentially, as long as 0<x/2< $\pi/2$ (although this is not an iff condition) , same for 3x; you want 3x to fall within an interval where the property holds. This is essentially a change of variable.