can anyone prove the identity of cos(2x) = 1 - tan ^2 x / 1 + tan ^2 x?

Answer 1

[1-tan^2(x)]/[1+tan^2(x)]

Write tan(x) as sin(x)/cos(x)

= [1-sin^2(x)/cos^2(x)]/ [1+sin^2(x)/cos^2(x)]

= {[cos^2(x)-sin^2(x)]/cos^2(x)}/{[cos^2(x)+sin^2(x)]/cos^2(x)}

= [cos^2(x)-sin^2(x)]/[cos^2(x)+sin^2(x)]

but [cos^2(x) + sin^2(x)] = 1 so

= [cos^2(x) - sin^2(x)]

Using the sum of angles formula for cosines (i.e., cos(x + y) = cos(x)*cos(y) - sin(x)*sin(y)) and letting y = x gets us that:

cos(2x) = cos(x)*cos(x) -sin(x)*sin(x) = cos^2(x) - sin^2(x)

QED

Answer 2

[1] .. cos 2x = (cos x)^2 - (sin x)^2

This looks a little bit like

[2] ... 1 = (cos x)^2 + (sin x)^2

Divide [1]/[2], we find

[3] ... cos 2x / 1 = [(cos x)^2 - (sin x)^2]/[(cos x)^2 + (sin x)^2]

Now divide all terms at the right side by cos x, and use the fact that (sin x)/(cos x) = tan x.


(As for the "disproving": tan (Pi/2) is not equal to zero, but infinity. In that case, the formula is undefined, but not contradictory.)

Answer 3

cos (2x) = (1 - tan^2 x) / (1 + tan^2 x)

= [1 - (sin^2 x / cos^2 x) ] / (sec^2 x) ----> by Pythagorean identity
= cos^2 x * [1 - (sin^2 x / cos^2 x) ]

=cos^2 x - sin^2 x
= cos (2x) -----> by the double-angle identity

Answer 4

Nope, but I can easily disprove it.

Take x = Pi/2

Then cos(2x)= -1

and 1 - tan ^2 2x / 1 + tan ^2 2x = 1/1 = 1 =/= -1



QED

edit:

oops, I did it for (1-tan^2 2x)/(1+tan^2 2x), big difference, but...

In which case, at odd ntuples of pi/2, we've got an undefined quantity, in which case we still *do not* have equality - we've got equivalence almost everywhere. There is a difference.