anonymous
  • anonymous
Q9:suppose that 0
Mathematics
jamiebookeater
  • jamiebookeater
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slaaibak
  • slaaibak
not sure how to do this, but I would use a comparison test.. maybe showing that first: tan xn = sin xn/cos xn = nsinxn / ncosxn then showing: [sin x is increasing on x e (0, pi/2) ] (n-1)(sin x(n-1)) + sin xn < nsinxn [cos x is decreasing on x e (0, pi/2) ] and (n-1)(cos x(n-1)) + cos xn > ncosxn Then [(n-1)(sin x(n-1)) + sin xn]/[(n-1)(cos x(n-1)) + cos xn] < n sin xn/ n cos xn Similar reasoning to show it's > tan x1. but ye, Just a suggestion, it's probably wrong. but ye use if you can..
anonymous
  • anonymous
what is a (n-1)sinx(n-1)?and how you use that?

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slaaibak
  • slaaibak
it's \[\sin x_{n-1}\] n-1 times... so \[\sin x_{n-1} +\sin x_{n-1} +\sin x_{n-1} +\sin x_{n-1} + ... +\sin x_{n-1}\]
anonymous
  • anonymous
I know !bot I want to solve: tanx1<(sinx1+sinx2+...+sinxn)/(cosx1+cosx2+...+cosxn)
slaaibak
  • slaaibak
well, you can't solve that. I'm saying you can use a comparison to prove it... Read my first message thoroughly
anonymous
  • anonymous
I see your first commend. But choosing those two terms in num and den confusing. However I accept your opinion for sin and cos.
anonymous
  • anonymous
\[[n.\sin x _{1} \le \sin x _{1}+\sin x _{2}+...\sin x _{n}\le n.\sin x _{n}\\] n.\cos x _{n} \le \cos x _{1}+\cos x _{2}+...\cos x _{n}\le n.\cos x _{1}\]\] as a known a a/C

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