Prove that $ an(2 heta) = frac{2 an( heta)}{1 - an^2( heta)} $ using the unit circle

Consider a point on the unit circle at an angle $ \theta $. Using the double-angle identities, we can write:

$ \sin(2\theta) = 2\sin(\theta)\cos(\theta) $

and

$ \cos(2\theta) = \cos^2(\theta) – \sin^2(\theta) $

Thus,

$ \tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta) – \sin^2(\theta)} $

Using the fact that $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $, let

$ t = \tan(\theta) $

Then

$ \sin(\theta) = \frac{t}{\sqrt{1 + t^2}} $

and

$ \cos(\theta) = \frac{1}{\sqrt{1 + t^2}} $

Substituting these into the double-angle formula gives:

$ \tan(2\theta) = \frac{2 \cdot \frac{t}{\sqrt{1 + t^2}} \cdot \frac{1}{\sqrt{1 + t^2}}}{\left(\frac{1}{\sqrt{1 + t^2}}\right)^2 – \left(\frac{t}{\sqrt{1 + t^2}}\right)^2} $

Which simplifies to:

$ \tan(2\theta) = \frac{2t}{1 – t^2} $

We start by using the double-angle identities for sine and cosine:

$ sin(2 heta) = 2sin( heta)cos( heta) $

and

$ cos(2 heta) = cos^2( heta) – sin^2( heta) $

Thus,

$ an(2 heta) = frac{sin(2 heta)}{cos(2 heta)} = frac{2sin( heta)cos( heta)}{cos^2( heta) – sin^2( heta)} $

Let $ t = an( heta) $, so

$ sin( heta) = frac{t}{sqrt{1 + t^2}} $

and

$ cos( heta) = frac{1}{sqrt{1 + t^2}} $

Substituting these into the double-angle formula, we get:

$ an(2 heta) = frac{2 cdot frac{t}{sqrt{1 + t^2}} cdot frac{1}{sqrt{1 + t^2}}}{left(frac{1}{sqrt{1 + t^2}}
ight)^2 – left(frac{t}{sqrt{1 + t^2}}
ight)^2} $

Simplifying, we find:

$ an(2 heta) = frac{2t}{1 – t^2} $

Start with the double

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