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Problem 15 – Solution

Posted in Solutions for Posted Problems by Lia on 06/19/2011

The problem:

Given:

{{sin(x + z)}/{sin(y - z)}} = m and  {{cos(x - z)}/{cos(y + z)}} = n

Show that the identity:

{{cos{2z}}/{cos(x - y)}} = {{m + n}/{mn + 1}}

is true.

Admission exam to Engineering,
Polytechnic Institute of Bucharest, Romania, 19xx
Our Translation.

We did not receive any good solution therefore we post

Our Solution:

Let’s start by replacing  {{}{m}{}} and {{}{n}{}} in {{m + n}/{mn + 1}}, with the given values:

{{m + n}/{mn + 1}} = {{{{sin(x + z)}/{sin(y - z)}} + {{cos(x - z)}/{cos(y + z)}}}/{({sin(x + z)}/{sin(y - z)})({cos(x - z)}/{cos(y + z)}) + 1}} = {{{{sin(x + z)}{cos(y + z)} + {sin(y - z)}{cos(x - z)}} /{{sin(y - z)}{cos(y + z)}}}/{{{sin(x + z)}{cos(x-z)} + {sin(y - z)}{cos(y + z)}} /{{sin(y - z)}{cos(y + z)}}}}

And:

[1]  {{m + n}/{mn + 1}} = {{{sin(x + z)}{cos(y + z)} + {sin(y - z)}{cos(x - z)}}/{{sin(x + z)}{cos(x-z)} + {sin(y - z)}{cos(y + z)}}}

Using the formula:

sin{alpha}cos{beta} = {{sin({alpha} + {beta}) + sin({alpha} - {beta})}/2}

we have:

a)  {sin(x + z)}{cos(y + z)} = {{{sin(x + z + y + z)} + {sin(x + z - y - z)}}/2} = {{{sin(x + y +2z)} + {sin(x - y)}}/2}

b) {sin(y - z)}{cos(x - z)} = {{{sin(y - z + x - z)} + {sin(y - z - x + z)}}/2} = {{{sin(x + y - 2z)} + {sin(y - x)}}/2}

c) {sin(x + z)}{cos(x - z)} = {{{sin(x + z + x - z)} + {sin(x + z - x + z)}}/2} = {{{sin(2x)} + {sin(2 z)}}/2}

d) {sin(y -z)}{cos(y + z)} = {{{sin(y - z + y + z)} + {sin(y - z - y - z)}}/2} = {{{sin(2y)} + {sin(-2 z)}}/2}

but {sin( - {alpha}) = - sin{alpha}} therefore {sin( - 2z) = - sin{2z}}

and {sin(y - z)}{cos(y + z)} = {{{sin(2y)} - {sin(2 z)}}/2}

By replacing the expressions  from a), b), c) and d) into [1] we have:

{{m + n}/{mn + 1}} = {{{sin(x + z)}{cos(y + z)} + {sin(y - x)}{cos(x - z)}}/{{sin(x + z)}{cos(x-z)} + {sin(y - z)}{cos(y + z)}}} = {{{{{sin(x + y +2z)} + {sin(x - y)}}/2} + {{{sin(x + y - 2z)} + {sin(y - x)}}/2}}/{{{{sin(2x)} + {sin(2z)}}/2} + {{{sin(2y)} - {sin(2 z)}}/2}}} =

{{{{sin(x + y +2z)} + {sin(x - y)}} + {{sin(x + y - 2z)} + {sin(y - x)}}}/{{{sin(2x)} + {sin(2z)}} + {{sin(2y)} - {sin(2 z)}}}} =

= {{{sin(x + y +2z)} + {sin(x - y)} + {sin(x + y - 2z)} - {sin(x - y)}}/{{sin(2x)} + {sin(2z)} + {sin(2y)} - {sin(2 z)}}}

And:

[2]   {{m + n}/{mn + 1}} = {{{sin(x + y +2z)} + {sin(x + y - 2z)}}/{{sin(2x)}+ {sin(2y)}}}

Using the formula:

sin{alpha} + sin{beta}  = 2 sin({{alpha} + {beta}}/2)cos({{alpha} - {beta}}/2)

the expression in [2] becomes:

{{{sin(x + y +2z)} + {sin(x + y - 2z)}}/{{sin(2x)} + {sin(2y)}}}={{{2sin({x + y + 2z + x + y - 2z}/2)cos({x + y + 2z - x -y + 2z}/2)}/{2sin({2x + 2y}/2)cos({2x - 2y}/2)}}} =

{{2sin{{2x + 2y}/2}cos{{4z}/2}}/{2sin{{2(x + y)}/2}cos{{2(x - y)}/2}}} = {{2sin(x + y)cos(2z)}/{2sin(x + y)cos(x - y)}} = {{cos{2z}}/{cos(x-y)}}

That’s means that:

{{m + n}/{mn + 1}} = {{cos{2z}}/{cos(x - y)}}

is true.