If #x_1#,#x_2#,#x_3#,......,#x_n# are #n# real numbers satisfying #0<x_1<x_2<...<x_n<pi/2#, how do you prove that #tanx_1 < (sinx_1+sinx_2+...+sinx_n)/ (cosx_1+cosx_2+...+cosx_n) < tanx_n#?

2 Answers
George C.
Aug 10, 2016

See explanation...

Explanation:

Here are the things I will assume are known:

#tan x = (sin x)/(cos x)#

In the interval #(0, pi/2)# we have:

#{ (sin x > 0), (cos x > 0), (tan x " is strictly monotonic increasing with " x) :}#

For brevity use the abbreviations:

#{ (s_k = sin x_k), (c_k = cos x_k), (t_k = tan x_k) :}#

Note that if #a, b, c, d > 0# then:

#a/b < c/d# if and only if #ad < bc#

So:

#s_1/c_1 < s_2/c_2 " " =>#

#s_1 c_2 < s_2 c_1 " " =>#

#s_1 c_2 + s_1 c_1 < s_2 c_1 + s_1 c_1 " " =>#

#s_1(c_1+c_2) < c_1(s_1+s_2) " " =>#

#s_1/c_1 < (s_1+s_2)/(c_1+c_2)#

and:

#s_1/c_1 < s_2/c_2 " " =>#

#s_1 c_2 < s_2 c_1 " " =>#

#s_1 c_2 + s_2 c_2 < s_2 c_1 + s_2 c_2 " " =>#

#(s_1+s_2)c_2 < (c_1+c_2)s_2 " " =>#

#(s_1+s_2)/(c_1+c_2) < s_2/c_2#

So we find:

#s_1/c_1 < (s_1+s_2)/(c_1+c_2) < s_2/c_2#

In general, if we have proved (for some #k < n#):

#(s_1+s_2+...+s_k)/(c_1+c_2+...+c_k) < s_k/c_k#

then since #s_k/c_k < s_(k+1)/c_(k+1)# we have:

#(s_1+s_2+...+s_k)/(c_1+c_2+...+c_k) < s_(k+1)/c_(k+1)#

and hence:

#((s_1+s_2+...+s_k))/((c_1+c_2+...+c_k)) < ((s_1+s_2+...+s_k)+s_(k+1))/((c_1+c_2+...+c_k)+c_(k+1))#

Applying this for each #k < n# in increasing order, we eventually get to:

#s_1/c_1 < (s_1+s_2+...+s_n)/(c_1+c_2+...+c_n)#

Similarly, if we have proved (for some #k > 1#):

#s_k/c_k < (s_k+s_(k+1)+...+s_n)/(c_k+c_(k+1)+...+c_n)#

then since #s_(k-1)/c_(k-1) < s_k/c_k# we have:

#s_(k-1)/c_(k-1) < (s_k+s_(k+1)+...+s_n)/(c_k+c_(k+1)+...+c_n)#

and hence:

#(s_(k-1)+(s_k+s_(k+1)+...+s_n))/(c_(k-1)+(c_k+c_(k+1)+...+c_n)) < (s_k+s_(k+1)+...+s_n)/(c_k+c_(k+1)+...+c_n)#

Applying this for each #k > 1# in decreasing order, we eventually get to:

#(s_1+s_2+...+s_n)/(c_1+c_2+...+c_n) < s_n/c_n#

So #tan x_1 = (sin x_1)/(cos x_1) < (sin x_1 + sin x_2 +...+ sin x_n)/(cos x_1 + cos x_2 +...+ cos x_n) < (sin x_n)/(cos x_n) = tan x_n#

Cesareo R.
Aug 10, 2016

See below

Explanation:

Given two real fractions such that

#x_1/(y_1) < x_2/(y_2)# then

#x_1/(y_1) < (x_1+x_2)/(y_1+y_2) < x_2/(y_2)# because

calling #lambda_1 = (x_1)/(y_1)# and #lambda_2 = (x_2)/(y_2)#

we have

#lambda_1 < (lambda_1 y_1 + lambda_2 y_2)/(y_1 + y_2) < lambda_2#

but #y_1/(y_1+y_2) + y_2/(y_1+y_2) = 1# so calling

#mu = y_2/(y_1+y_2)# we have

#lambda_1 < (1-mu)lambda_1 + mu lambda_2 < lambda_2#

or

#lambda_1 < lambda_1 + mu(lambda_2-lambda_1) < lambda_2# This is true for #0 < mu < 1#

Now generalizing, if we have

#x_1/(y_1) < x_2/(y_2) < cdots < x_n/(y_n)# then

#x_1/(y_1) < (sum_{i = 1}^n x_i)/(sum_{i = 1}^n y_i) < x_n/(y_n)#

We know that if

#0 < x_1 < x_2 < cdots < x_n < pi/2# then
#tan(x_1) < tan(x_2) < cdots < tan(x_n)# because

#tan(x)# is strict monotonic increasing in this domain

Related questions
See all questions in Square Roots and Irrational Numbers
Impact of this question
1863 views around the world
You can reuse this answer
Creative Commons License