# Prove? cotx = sin5x+sin7x / cos5x-cos7x

Mar 21, 2018

See the proof below

#### Explanation:

Reminder

$\sin a + \sin b = 2 \sin \left(\frac{a + b}{2}\right) \cos \left(\frac{a - b}{2}\right)$

$\cos a - \cos b = - 2 \sin \left(\frac{a + b}{2}\right) \sin \left(\frac{a - b}{2}\right)$

Therefore,

$R H S = \frac{\sin 5 x + \sin 7 x}{\cos 5 x - \cos 7 x}$

$= \frac{2 \sin \left(\frac{5 x + 7 x}{2}\right) \cos \left(\frac{5 x - 7 x}{2}\right)}{- 2 \sin \left(\frac{5 x + 7 x}{2}\right) \sin \left(\frac{5 x - 7 x}{2}\right)}$

$= \frac{\cos \left(\frac{5 x - 7 x}{2}\right)}{- \sin \left(\frac{5 x - 7 x}{2}\right)}$

$= \cos \frac{x}{- \left(- \sin x\right)}$

$= \cot x$

$= L H S$

$Q E D$

Mar 21, 2018

#### 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,
$\cot x = \frac{\sin 5 x + \sin 7 x}{\cos 5 x - \cos 7 x}$

$R H S = \frac{\sin 5 x + \sin 7 x}{\cos 5 x - \cos 7 x}$

Using (1) and (2),

$= \frac{\cancel{2} \cancel{\sin} \left(\frac{5 x + 7 x}{2}\right) \cos \left(\frac{5 x - 7 x}{2}\right)}{- \cancel{2} \cancel{\sin} \left(\frac{5 x + 7 x}{2}\right) \sin \left(\frac{5 x - 7 x}{2}\right)}$

$= - \cos \frac{\frac{5 x - 7 x}{2}}{\sin} \left(\frac{5 x - 7 x}{2}\right)$

$= - \cos \frac{\frac{- 2 x}{2}}{\sin} \left(\frac{- 2 x}{2}\right)$

$= - \cos \frac{- x}{\sin} \left(- x\right)$

=-cosx/(-sinx

$= \cot x$