Question

Q9:suppose that 0 13 years ago

Answer 1

*

Answer 2

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..

Answer 3

what is a (n-1)sinx(n-1)?and how you use that?

Answer 4

it's
\[\sin x_{n-1}\]

n-1 times... so

\[\sin x_{n-1} +\sin x_{n-1} +\sin x_{n-1} +\sin x_{n-1} + ... +\sin x_{n-1}\]

Answer 5

I know !bot I want to solve:
tanx1<(sinx1+sinx2+...+sinxn)/(cosx1+cosx2+...+cosxn)<tanxn

Answer 6

well, you can't solve that. I'm saying you can use a comparison to prove it...

Read my first message thoroughly

Answer 7

I see your first commend.
But choosing those two terms in num and den confusing.
However I accept your opinion for sin and cos.

Answer 8

\[[n.\sin x _{1} \le \sin x _{1}+\sin x _{2}+...\sin x _{n}\le n.\sin x _{n}\\]
n.\cos x _{n} \le \cos x _{1}+\cos x _{2}+...\cos x _{n}\le n.\cos x _{1}\]\]
as a known a<b<c ,A<B<C => a/C<b/B<c/A
so \[\tan x _{1} \le (\sin x _{1}+\sin x _{2}+...\sin x _{n})/(\cos x _{1}+\cos x _{2}+...\cos x _{n}) \le \tan x _{n}\]