Prove \(\displaystyle{{\tan{{A}}}_{{1}}{ < }}{\frac{{{\sum_{{{i}={1}}}^{{n}}}{\sin{{A}}}_{{1}}}}{{{\sum_{{{i}={1}}}^{{n}}}{\cos{{A}}}_{{1}}}}}{ < }{{\tan{{A}}}_{{n}}}\) for

kembdumatxf

kembdumatxf

Answered question

2022-04-01

Prove tanA1<i=1nsinA1i=1ncosA1<tanAn for 0<A1<A2<<An<π2

Answer & Explanation

exinnaemekswr1k

exinnaemekswr1k

Beginner2022-04-02Added 10 answers

This doesn't need any more trigonometry than the remark you've already made - what's actually going on is that if x1<x2<<xn and y1>y2>>yn are positive reals,
x1y1<x1+x2++xny1+y2++yn<xnyn
Try to prove this inequality (hint: a lower bound for the top of the fraction is nx1).
Jermaine Lam

Jermaine Lam

Beginner2022-04-03Added 11 answers

0<A1<A2<<An<π2
0<sinA1<sinA2<<sinAn<1
1>cosA1>cosA2>>cosAn>0
sinA1+sinA2++sinAncosA1+cosA2++cosAn>sinA1+sinA1++sinA1cosA1+cosA2++cosAn>sinA1+sinA1++sinA1cosA1+cosA1++cosA1=nsinA1ncosA1=tanA1
The other inequality involving tanAn is similar.

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