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How do you prove #tan(x) = sin(2x)/[1+cos(2x)]#?

Answer 1

Use the trig identities: #sin 2x = 2sin x.cos x# #(1 + cos 2x) = 2cos^2 x# Right side: #(sin 2x)/(1 + cos 2x) = (2sin x.cos x)/(2cos^2 x) = sin x/(cos x) = tan x.#

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Answer 2

Adam Aguirre

To prove ( \tan(x) = \frac{\sin(2x)}{1 + \cos(2x)} ), we'll use trigonometric identities.

Starting with the right-hand side: [ \frac{\sin(2x)}{1 + \cos(2x)} ]

Using the double angle identities: [ \sin(2x) = 2\sin(x)\cos(x) ] [ \cos(2x) = \cos^2(x) - \sin^2(x) ]

Substituting these into the expression: [ \frac{2\sin(x)\cos(x)}{1 + (\cos^2(x) - \sin^2(x))} ]

Using the Pythagorean identity: [ \cos^2(x) + \sin^2(x) = 1 ]

Substitute this identity into the expression: [ \frac{2\sin(x)\cos(x)}{1 + \cos^2(x) - \sin^2(x)} ]

[ \frac{2\sin(x)\cos(x)}{1 + \cos^2(x) - (1 - \cos^2(x))} ]

[ \frac{2\sin(x)\cos(x)}{1 + \cos^2(x) - 1 + \cos^2(x)} ]

[ \frac{2\sin(x)\cos(x)}{2\cos^2(x)} ]

[ \frac{\sin(x)}{\cos(x)} ]

[ \tan(x) ]

Hence, ( \tan(x) = \frac{\sin(2x)}{1 + \cos(2x)} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.