Prove? cotx = sin5x+sin7x / cos5x-cos7x

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Prove? cotx = sin5x+sin7x / cos5x-cos7x

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Answer 1 · 2018-03-21T07:07:06

See the proof below

Explanation:

Reminder

$sin a + sin b = 2 sin (\frac{a + b}{2}) cos (\frac{a - b}{2})$ $sina+sinb=2sin((a+b)/2)cos((a-b)/2)$

$cos a - cos b = - 2 sin (\frac{a + b}{2}) sin (\frac{a - b}{2})$ $cosa-cosb=-2sin((a+b)/2)sin((a-b)/2)$

Therefore,

$R H S = \frac{sin 5 x + sin 7 x}{cos 5 x - cos 7 x}$ $RHS=(sin5x+sin7x)/(cos5x-cos7x)$

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

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

$= \frac{cos x}{- (- sin x)}$ $=cosx/(-(-sinx))$

$= cot x$ $=cotx$

$= L H S$ $=LHS$

$Q E D$ $QED$

Answer 2 · 2018-03-21T07:16:31

Please see below.

Explanation:

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

$(2) cos C - cos D = - 2 sin (\frac{C + D}{2}) sin (\frac{C - D}{2})$ $color(red)((2)cosC-cosD=-2sin((C+D)/2)sin((C-D)/2)$
Here,
$cot x = \frac{sin 5 x + sin 7 x}{cos 5 x - cos 7 x}$ $cotx=(sin5x+sin7x)/(cos5x-cos7x)$

$R H S = \frac{sin 5 x + sin 7 x}{cos 5 x - cos 7 x}$ $RHS=(sin5x+sin7x)/(cos5x-cos7x)$

Using (1) and (2),

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

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

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

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

$=-cosx/(-sinx$

$=cotx$

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Prove? cotx = sin5x+sin7x / cos5x-cos7x

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