3 years, 2 months ago
Math whizzes: prove sin x = cos x ( cot x - 2 cot 2x )
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Ok I'm gonna do this as simply as I can.
I found a proof on another website that:
cot 2x = (1/2)(cot x - tan x)
Here is the proof:
(1) cot 2x = 1/tan(2x) = cos(2x)/sin(2x)
(2) cos(2x) = cos^2(x) - sin^2(x)
(3) sin(2x) = 2(sin x)(cos x)
(4) cos(2x)/sin(2x) = (cos^2(x) - sin^2(x))/2(sin x)(cos x) =
= (1/2)(cos(x)/sin(x) - sin(x)/cos(x)) = (1/2)(cot(x) - tan(x))
It explains this more clearly here:
http://mathrefresher.blogspot.com/2008/01/cot-2x-12-tan-x-cot-x.html
So now we have:
cot 2x = (1/2)(cot x - tan x)
tan(x) = sin(x)/cos(x) (by definition)
And the equation we are proving is:
sin x = cos x ( cot x - 2 cot 2x )
So to substitute for cot 2x
sin x = cos x ( cot x - 2 {(1/2)(cot x - tan x)}
sin x = cos x ( cot x - cot x + tan x)
sin x = cos x (tan x)
sin x = cos x (sin x / cos x)
sin x = sin x
I'm really bad at trying to explain math on the internet, but I hope this is understandable.
I found a proof on another website that:
cot 2x = (1/2)(cot x - tan x)
Here is the proof:
(1) cot 2x = 1/tan(2x) = cos(2x)/sin(2x)
(2) cos(2x) = cos^2(x) - sin^2(x)
(3) sin(2x) = 2(sin x)(cos x)
(4) cos(2x)/sin(2x) = (cos^2(x) - sin^2(x))/2(sin x)(cos x) =
= (1/2)(cos(x)/sin(x) - sin(x)/cos(x)) = (1/2)(cot(x) - tan(x))
It explains this more clearly here:
http://mathrefresher.blogspot.com/2008/01/cot-2x-12-tan-x-cot-x.html
So now we have:
cot 2x = (1/2)(cot x - tan x)
tan(x) = sin(x)/cos(x) (by definition)
And the equation we are proving is:
sin x = cos x ( cot x - 2 cot 2x )
So to substitute for cot 2x
sin x = cos x ( cot x - 2 {(1/2)(cot x - tan x)}
sin x = cos x ( cot x - cot x + tan x)
sin x = cos x (tan x)
sin x = cos x (sin x / cos x)
sin x = sin x
I'm really bad at trying to explain math on the internet, but I hope this is understandable.
You can leave an optional "tip" with Mahalo's virtual currency, Mahalo Dollars. If you are asking a difficult question that might require some research, or if you'd like a wide variety of feedback, a higher tip often leads to more answers to your question.
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