Prove? cotx = sin5x+sin7x / cos5x-cos7x

2 Answers
Mar 21, 2018

See the proof below

Explanation:

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

Mar 21, 2018

Please see below.

Explanation:

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