anonymous
  • anonymous
cosec x= 1/2 (x)+1 has root 0
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Remember -1 <= sin x <= 1 Therefore -1 <= cosec x <=1 Now try :)
anonymous
  • anonymous
Still not sure..
anonymous
  • anonymous
One more question. This is your question , right \[cosec x= {1 \over 2 }(x)+1\]

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anonymous
  • anonymous
Yes
anonymous
  • anonymous
I think i solved it
anonymous
  • anonymous
OK. Just put instead of cosec x , 1/2(x) +1
anonymous
  • anonymous
Ok @NotSObright , let's see what you have done :P
anonymous
  • anonymous
Say y1=1/2x+1 y2=csc(x) y1(0)=1 y1(Pi/2)=Pi/4+1 y2(0)=infinty y2(Pi/2)=1 Since it can be proven using differentiation y1 is inceasing and y2 is decreasing on 0,pi/2 hence in order to attain boundary values graphs must cross each other
anonymous
  • anonymous
So what about the 0.5 and 1?
anonymous
  • anonymous
sorry i didn't see that u must show at 0.5 y1y2 u might wanna use a calculator
anonymous
  • anonymous
But how?
anonymous
  • anonymous
|dw:1336990897568:dw|
anonymous
  • anonymous
in calculation?
anonymous
  • anonymous
it is in calculation u need to calculate values at o.5 and 1 to show the inequality
anonymous
  • anonymous
So, how exactly?
anonymous
  • anonymous
i have told everything u need go through the conversation and try to use ur intelligence
anonymous
  • anonymous
hmm. ok..

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