How to see that $sin^2left(frac{sqrt{x}}{2}right)leleft(frac{sqrt{x}}{2}right)^2$?
My calculusbook simply states that:
$$sin^2left(frac{sqrt{x}}{2}right)leleft(frac{sqrt{x}}{2}right)^2$$
...but I don't immediately see why this is true.
What (probably) simple trick am I missing? Apparently it is "obvious"...
algebra-precalculus
asked Nov 26 at 20:02
GambitSquared
1,1581137
add a comment |
My calculusbook simply states that:
$$sin^2left(frac{sqrt{x}}{2}right)leleft(frac{sqrt{x}}{2}right)^2$$
...but I don't immediately see why this is true.
What (probably) simple trick am I missing? Apparently it is "obvious"...
algebra-precalculus
asked Nov 26 at 20:02
GambitSquared
1,1581137
Do you know how to show that $|sin x| < |x|$?
– Connor Harris
Nov 26 at 20:10
@ConnorHarris Well, at least I know that to be true for a fact. Can we work from there to reach the above conclusion?
– GambitSquared
Nov 26 at 20:11
add a comment |
My calculusbook simply states that:
$$sin^2left(frac{sqrt{x}}{2}right)leleft(frac{sqrt{x}}{2}right)^2$$
...but I don't immediately see why this is true.
What (probably) simple trick am I missing? Apparently it is "obvious"...
algebra-precalculus
asked Nov 26 at 20:02
GambitSquared
1,1581137
My calculusbook simply states that:
$$sin^2left(frac{sqrt{x}}{2}right)leleft(frac{sqrt{x}}{2}right)^2$$
...but I don't immediately see why this is true.
What (probably) simple trick am I missing? Apparently it is "obvious"...
algebra-precalculus
algebra-precalculus
asked Nov 26 at 20:02
GambitSquared
1,1581137
asked Nov 26 at 20:02
GambitSquared
1,1581137
edited Nov 26 at 20:13
asked Nov 26 at 20:02
GambitSquared
1,1581137
asked Nov 26 at 20:02
GambitSquared
1,1581137
asked Nov 26 at 20:02
GambitSquared
1,1581137
1,1581137
Do you know how to show that $|sin x| < |x|$?
– Connor Harris
Nov 26 at 20:10
@ConnorHarris Well, at least I know that to be true for a fact. Can we work from there to reach the above conclusion?
– GambitSquared
Nov 26 at 20:11
add a comment |
Do you know how to show that $|sin x| < |x|$?
– Connor Harris
Nov 26 at 20:10
@ConnorHarris Well, at least I know that to be true for a fact. Can we work from there to reach the above conclusion?
– GambitSquared
Nov 26 at 20:11
Do you know how to show that $|sin x| < |x|$?
– Connor Harris
Nov 26 at 20:10
Do you know how to show that $|sin x| < |x|$?
– Connor Harris
Nov 26 at 20:10
@ConnorHarris Well, at least I know that to be true for a fact. Can we work from there to reach the above conclusion?
– GambitSquared
Nov 26 at 20:11
@ConnorHarris Well, at least I know that to be true for a fact. Can we work from there to reach the above conclusion?
– GambitSquared
Nov 26 at 20:11
add a comment |
2 Answers
2
active
oldest
votes
By MVT
$$sin(X)-sin(0)=Xcos(c)$$
thus
$$|sin(X)|le |X|$$
and
$$sin^2(X)le X^2$$
now apply to
$$X=frac{sqrt{x}}{2}$$
answered Nov 26 at 20:15
hamam_Abdallah
37.9k21634
add a comment |
This will follow if you show that $sin(t) leq t$ for all $tgeq 0$: To do that, you can try to show that if $f(0) = g(0)$ and $f'(t) leq g'(t)$ for all $tgeq 0$, then $f(t) leq g(t)$ for all $tgeq 0$.
answered Nov 26 at 20:08
user25959
1,563816
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
By MVT
$$sin(X)-sin(0)=Xcos(c)$$
thus
$$|sin(X)|le |X|$$
and
$$sin^2(X)le X^2$$
now apply to
$$X=frac{sqrt{x}}{2}$$
answered Nov 26 at 20:15
hamam_Abdallah
37.9k21634
add a comment |
By MVT
$$sin(X)-sin(0)=Xcos(c)$$
thus
$$|sin(X)|le |X|$$
and
$$sin^2(X)le X^2$$
now apply to
$$X=frac{sqrt{x}}{2}$$
answered Nov 26 at 20:15
hamam_Abdallah
37.9k21634
add a comment |
By MVT
$$sin(X)-sin(0)=Xcos(c)$$
thus
$$|sin(X)|le |X|$$
and
$$sin^2(X)le X^2$$
now apply to
$$X=frac{sqrt{x}}{2}$$
answered Nov 26 at 20:15
hamam_Abdallah
37.9k21634
By MVT
$$sin(X)-sin(0)=Xcos(c)$$
thus
$$|sin(X)|le |X|$$
and
$$sin^2(X)le X^2$$
now apply to
$$X=frac{sqrt{x}}{2}$$
answered Nov 26 at 20:15
hamam_Abdallah
37.9k21634
answered Nov 26 at 20:15
hamam_Abdallah
37.9k21634
answered Nov 26 at 20:15
hamam_Abdallah
37.9k21634
answered Nov 26 at 20:15
hamam_Abdallah
37.9k21634
37.9k21634
add a comment |
add a comment |
This will follow if you show that $sin(t) leq t$ for all $tgeq 0$: To do that, you can try to show that if $f(0) = g(0)$ and $f'(t) leq g'(t)$ for all $tgeq 0$, then $f(t) leq g(t)$ for all $tgeq 0$.
answered Nov 26 at 20:08
user25959
1,563816
add a comment |
This will follow if you show that $sin(t) leq t$ for all $tgeq 0$: To do that, you can try to show that if $f(0) = g(0)$ and $f'(t) leq g'(t)$ for all $tgeq 0$, then $f(t) leq g(t)$ for all $tgeq 0$.
answered Nov 26 at 20:08
user25959
1,563816
add a comment |
This will follow if you show that $sin(t) leq t$ for all $tgeq 0$: To do that, you can try to show that if $f(0) = g(0)$ and $f'(t) leq g'(t)$ for all $tgeq 0$, then $f(t) leq g(t)$ for all $tgeq 0$.
answered Nov 26 at 20:08
user25959
1,563816
This will follow if you show that $sin(t) leq t$ for all $tgeq 0$: To do that, you can try to show that if $f(0) = g(0)$ and $f'(t) leq g'(t)$ for all $tgeq 0$, then $f(t) leq g(t)$ for all $tgeq 0$.
answered Nov 26 at 20:08
user25959
1,563816
answered Nov 26 at 20:08
user25959
1,563816
answered Nov 26 at 20:08
user25959
1,563816
answered Nov 26 at 20:08
user25959
1,563816
1,563816
add a comment |
add a comment |
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Do you know how to show that $|sin x| < |x|$?
– Connor Harris
Nov 26 at 20:10
@ConnorHarris Well, at least I know that to be true for a fact. Can we work from there to reach the above conclusion?
– GambitSquared
Nov 26 at 20:11