Prove? cotx = sin5x+sin7x / cos5x-cos7x

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Answer 1 · 2018-03-21T07:07:06

See the proof below

Reminder

$sina+sinb=2sin((a+b)/2)cos((a-b)/2)$

$cosa-cosb=-2sin((a+b)/2)sin((a-b)/2)$

Therefore,

$RHS=(sin5x+sin7x)/(cos5x-cos7x)$

$=(2sin((5x+7x)/2)cos((5x-7x)/2))/(-2sin((5x+7x)/2)sin((5x-7x)/2))$

$=(cos((5x-7x)/2))/(-sin((5x-7x)/2))$

$=cosx/(-(-sinx))$

$=cotx$

$=LHS$

$QED$

Answer 2 · 2018-03-21T07:16:31

Please see below.

WE have,
$color(red)((1)sinC+sinD=2sin((C+D)/2)cos((C-D)/2)$

$color(red)((2)cosC-cosD=-2sin((C+D)/2)sin((C-D)/2)$
Here,
$cotx=(sin5x+sin7x)/(cos5x-cos7x)$

$RHS=(sin5x+sin7x)/(cos5x-cos7x)$

Using (1) and (2),

$=(cancel(2)cancelsin((5x+7x)/2)cos((5x-7x)/2))/(-cancel2cancelsin((5x+7x)/2)sin((5x-7x)/2))$

$=-cos((5x-7x)/2)/sin((5x-7x)/2)$

$=-cos((-2x)/2)/sin((-2x)/2)$

$=-cos(-x)/sin(-x)$

$=-cosx/(-sinx$

$=cotx$