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Mathematics 21 Online
OpenStudy (cybershadow):

I was going through one question and found out one inequality that For x<0 , Sinx>x But i didnt get it why. Can anyone help? :=|

OpenStudy (lastdaywork):

We all know that - sin|x| < |x| multiply by -1 on both side sin(- |x| ) > (- |x| ) & the rest is obvious :)

OpenStudy (cybershadow):

Oh ok , GOt it swimming through my neurons :D Thanks a Lot!

OpenStudy (lastdaywork):

:)

OpenStudy (cybershadow):

oh wait lol now i started the question again and got confused again , how is Sin|x| <|x| ,@LastDayWork

OpenStudy (lastdaywork):

There is a longer and more elaborate proof; but I personally prefer this one - sinx = x (for x = 0) now y=x has a constant slope (=1) but the slope of y=sinx is decreasing (from 1) Hence, for all x >0 ; x > sinx Tell me if that doesn't make sense & I'll give you another proof.

OpenStudy (cybershadow):

slope of y = sinx is decreasing ? i thought it was increasing and then decreasing? :S

OpenStudy (lastdaywork):

Let f(x) = sinx f'(x) = cosx Now what do you think about the slope of f(x) ??

OpenStudy (cybershadow):

slope is cos x

OpenStudy (cybershadow):

right?

OpenStudy (cybershadow):

now it has maximum value at cosx=0 i.e. at x = Pi/2 , now thats all i got lol

OpenStudy (lastdaywork):

Slope of f(x) = f'(x) = cos x Now; at x = 0 ; y = sin x has slope = cos 0 = 1 For x > 0 ; sin x has slope = cos x < 1

OpenStudy (cybershadow):

oh right

OpenStudy (cybershadow):

Guess i got it lol thanx A LOT!!

OpenStudy (lastdaywork):

:)

OpenStudy (cybershadow):

thx

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