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
Aug 10, 2016

Answer:

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#

Aug 10, 2016

Answer:

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