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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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  • $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$ – Julian Kuelshammer Jan 13 '13 at 19:53
  • $\begingroup$ 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/… $\endgroup$ – Michael Hardy Jan 13 '13 at 20:10
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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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    $\begingroup$ @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. $\endgroup$ – Mark Bennet Jan 13 '13 at 20:07
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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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