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Prove the following: $$\tan \left(\frac{x}{2}\right) = \frac{1 + \sin (x) - \cos (x)}{1 + \sin (x) + \cos (x)}$$

I was unable to find any proofs of the above formula online. Thanks!

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My answer below may be the simplest, but I don't do it from scratch, but rather I deduce it from two other well known tangent half-angle formulas. There's also this somewhat related question: math.stackexchange.com/questions/113451/… –  Michael Hardy Jan 13 '13 at 20:10

4 Answers 4

Hints:

  • $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$
  • $1 - \cos x = 2 \sin^2 \frac{x}{2}$
  • $1 + \cos x = 2 \cos^2 \frac{x}{2}$
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There are several ways to proceed, apart from what Ayman gave

Approach 1: If you know the tangent t-formulas: For $t = \tan \frac {\theta} {2}$,

$$ \sin \theta = \frac {2t}{1+t^2}, \cos \theta = \frac {1-t^2}{1+t^2}$$

Substitute these into the equation and simplify.

Approach 2: (my preference)If you know that $\tan \frac {\theta}{2} = \frac {\sin \theta}{1+\cos \theta} = \frac { 1- \cos \theta}{\sin \theta}$, then by the equivalence class of fractions, we can sum across the numerator and denominator to get

$$\tan \frac {\theta}{2} = \frac {\sin \theta + 1 - \cos \theta}{ 1 + \cos \theta + \sin \theta}$$

Of course, you could also take the difference across the numerator and denominator to get

$$\tan \frac {\theta}{2} = \frac {\sin \theta - 1 + \cos \theta}{ 1 + \cos \theta - \sin \theta}$$

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@sdf You aren't adding fractions - try it with $\cfrac 1 2$ and $\cfrac 2 4$ and similar examples to get an idea why it might work - it only works like this when the fractions are equal. –  Mark Bennet Jan 13 '13 at 20:07

One well known tangent half-angle formula says $$ \tan\frac x2 = \frac{\sin x}{1+\cos x}. $$ Another well known tangent half-angle formula says: $$ \tan\frac x2 = \frac{1-\cos x}{\sin x}. $$

If two fractions $\dfrac AB$ and $\dfrac CD$ are equal, then their common value is also equal to $\dfrac{A+C}{B+D}$, so there you have it.

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You could always use Euler's formula, setting $$ z = e^{i x} $$ which gives for the left $$ \tan\left(\frac{x}{2}\right) = \frac{1}{i} \frac{z^{1/2}-z^{-1/2}}{z^{1/2}+z^{-1/2}} = \frac{1}{i} \frac{z-1}{z+1}$$ and for the right $$ \frac{1+\sin(x)-\cos(x)}{1+\sin(x)+\cos(x)} = \frac{1+ 1/(2i) z - 1/(2i) 1/z - (1/2) z - (1/2) 1/z} {1+ 1/(2i) z - 1/(2i) 1/z + (1/2) z + (1/2) 1/z} = \frac{z + 1/(2i) z^2 - 1/(2i) - (1/2) z^2 - (1/2)} {z + 1/(2i) z^2 - 1/(2i) + (1/2) z^2 + (1/2)} $$ which is $$ \frac{-1/2(1+i)(z-1)(z+i)}{1/2(1-i)(z+1)(z+i)} = -\frac{1+i}{1-i} \frac{z-1}{z+1} = \frac{1}{i} \frac{z-1}{z+1}.$$ While this may not be the most intuitive approach it does show that these trigonometric identities are amenable to automated theorem proving -- just make the substitution from Euler's formula, factorize LHS and RHS, done.

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