13
$\begingroup$

Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end.

Prove that, $$ \dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2} $$

$\endgroup$
1
  • 1
    $\begingroup$ The fact that you need to use half angle (or equivalently double angle) identities is built into the shape of the problem. The simple formulas express $\sin(2x)$, $\cos(2x)$, $\tan(2x)$ in terms of trigonometric functions of $x$, so the mechanical way to proceed is to let $x=A/2$. Then $A=2x$. Now work on the left-hand side. $\endgroup$ Commented Jul 7, 2011 at 13:49

8 Answers 8

18
$\begingroup$

Now that OP has understood how to prove this, here is a geometric proof for certain angles, just for fun :-)

Consider the figure:

enter image description here

$\displaystyle \triangle ABC$ is a right angled triangle with the right angle being at $\displaystyle C$.

$\displaystyle \angle{CAB} = A$ and $\displaystyle AB = 1$ and thus $\displaystyle BC = \sin A$ and $\displaystyle AC = \cos A$.

Now $\displaystyle AO$ is the angular bisector of $\displaystyle \angle{CAB}$. We select $\displaystyle O$ so that $\displaystyle O$ is the in-center (point of intersection of angular bisectors of a triangle). Let the in-radius $\displaystyle OD$ be $\displaystyle r$.

Now $\displaystyle BE = BF$ and $\displaystyle AE = AD$ and adding gives us $\displaystyle BF + AD = AB = 1$

Now $\displaystyle BF = BC - FC =BC - OD$ (as $\displaystyle ODCF$ is a square).

Thus $\displaystyle BF = \sin A - r$. Similarly $\displaystyle AD = \cos A - r$.

Thus $\displaystyle \sin A + \cos A - 2r = 1$.

Using $\displaystyle \triangle ADO$, $\displaystyle \tan \frac{A}{2} = \frac{OD}{AD} = \frac{r}{\cos A - r}$

Since $\displaystyle 2r = \sin A + \cos A -1 $ we get

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

It is easy to verify that

$$ \frac{\sin A + \cos A - 1}{\cos A - \sin A + 1} = \frac{\sin A - \cos A + 1}{\cos A + \sin A + 1}$$



Incidently, the fact that in a right triangle (hypotenuse $c$), the in-radius is given by $ a + b -c = 2r$ and also by $r = \frac{\triangle}{s}$ ($\triangle$ is the area, $s$ is the semi-perimeter) can be used to prove the pythagoras theorem ($a^2 + b^2 = c^2$)!

$\endgroup$
4
  • $\begingroup$ Awesome, how did u think of constructing such a triangle. $\endgroup$
    – user9413
    Commented Jul 7, 2011 at 19:56
  • $\begingroup$ Great Yaar... simply superb $\endgroup$
    – user9413
    Commented Jul 7, 2011 at 20:14
  • 1
    $\begingroup$ If I could upvote this a dozen times, I would. Gorgeous! $\endgroup$ Commented Jul 7, 2011 at 20:41
  • $\begingroup$ Wow!, This was a somewhat difficult proof for me, you've proved it geometrically! $\endgroup$
    – mathguy80
    Commented Jul 8, 2011 at 4:29
16
$\begingroup$

$\textbf{Method 1.}$ You have

\begin{align*} \frac{1+ \sin{A} - \cos{A}}{1+\sin{A} + \cos{A}} &= \frac{ \cos^{2}\frac{A}{2} + \sin^{2}\frac{A}{2} + 2\cdot \sin\frac{A}{2}\cdot\cos\frac{A}{2} - \Bigl(\cos^{2}\frac{A}{2} - \sin^{2}\frac{A}{2}\Bigr)}{\cos^{2}\frac{A}{2} + \sin^{2}\frac{A}{2} + 2\cdot \sin\frac{A}{2}\cdot\cos\frac{A}{2} + \Bigl(\cos^{2}\frac{A}{2} -\sin^{2}\frac{A}{2}\Bigr)} \\ &=\frac{ \Bigl(\cos\frac{A}{2} + \sin\frac{A}{2}\Bigr)^{2} - \Bigl(\cos\frac{A}{2}+\sin\frac{A}{2}\Bigr) \cdot \Bigl(\cos\frac{A}{2} - \sin\frac{A}{2}\Bigr)}{\Bigl(\cos\frac{A}{2} + \sin\frac{A}{2}\Bigr)^{2} + \Bigl(\cos\frac{A}{2}+\sin\frac{A}{2}\Bigr) \cdot \Bigl(\cos\frac{A}{2} - \sin\frac{A}{2}\Bigr)} \\ &= \frac{ \cos\frac{A}{2} + \sin\frac{A}{2} - \cos\frac{A}{2} + \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2} + \cos\frac{A}{2} -\sin\frac{A}{2}} \qquad \Bigl[ \text{cancelling}\ \Bigl(\small\cos\frac{A}{2} +\sin\frac{A}{2}\Bigr) \ \Bigr] \\ &= \tan\frac{A}{2} \end{align*}

$\textbf{Method 2.}$ Here is another way of seeing this. By using something called Componendo and Dividendo.

Let $$\frac{1+\sin{A} -\cos{A}}{1+\sin{A} + \cos{A}} = \frac{k}{1}$$ Now applying componendo and dividendo we get $$\frac{1+ \sin{A}}{-\cos{A}} = \frac{k+1}{k-1} \Longrightarrow \frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{-\cos\frac{A}{2}+\sin\frac{A}{2}} = \frac{k+1}{k-1} \qquad (1)$$

Again using componendo dividendo on $(1)$ we get $$\frac{\cos\frac{A}{2}+\sin\frac{A}{2} -\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2} +\cos\frac{A}{2}-\sin\frac{A}{2}} = \frac{k+1 +k-1}{k+1-k+1} \Longrightarrow k=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}=\tan\frac{A}{2}$$

$\textbf{Method 3.}$ Another way of looking into this would be, \begin{align*} \frac{1+\sin{A}-\cos{A}}{1+\sin{A}+\cos{A}} =\frac{(1-\cos{A})+\sin{A}}{(1+\cos{A}) + \sin{A}} &= \frac{ 2\:\sin^{2}\frac{A}{2} + 2\cdot\sin\frac{A}{2}\cdot\cos\frac{A}{2}}{2\:\cos^{2}\frac{A}{2} + 2 \cdot\sin\frac{A}{2}\cdot \cos\frac{A}{2}} \\ &= \frac{2\:\sin\frac{A}{2}}{2\:\cos\frac{A}{2}} \cdot \Biggl(\frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}}\Biggr) \\ &= \tan\frac{A}{2} \end{align*}

$\textbf{Method 4.}$ My younger brother mentioned about this method. Multiplying numerator and deominator by $(1+\sin{A}-\cos{A})$ we get \begin{align*} \frac{1+\sin{A}-\cos{A}}{1+\sin{A}+\cos{A}} \times \frac{1+\sin{A}-\cos{A}}{1+\sin{A}-\cos{A}} &= \frac{(1+\sin{A}-\cos{A})^{2}}{(1+\sin{A})^{2}-\cos^{2}{A}} \\ &= \frac{2 + 2\: \sin{A} -2\cdot\sin{A}\cdot \cos{A} -2\cos{A}}{2\: \sin{A} + 2\sin^{2}{A}} \\ &= \frac{(2-2\cos{A}) \cdot (1+\sin{A})}{2\:\sin{A} \cdot (1+\sin{A})} \\ &=\frac{1-\cos{A}}{\sin{A}} = \frac{2\: \sin^{2}\frac{A}{2}}{2\cdot \sin\frac{A}{2} \cdot \cos\frac{A}{2}} \\ &=\tan\frac{A}{2} \end{align*}

Oh once again when I looked at the question I realize that you attempted this method. Hopefully now it's clear.

$\endgroup$
6
  • $\begingroup$ Ah, the old $1=\cos^2(A/2)+\sin^2(A/2)$ trick. $\endgroup$ Commented Jul 7, 2011 at 13:49
  • 1
    $\begingroup$ @Gerry: Yes, may be not too old for me :) $\endgroup$
    – user9413
    Commented Jul 7, 2011 at 13:51
  • $\begingroup$ Nice! I ended up with something similar to your first example. Also Thanks for the componendo and dividendo solution! $\endgroup$
    – mathguy80
    Commented Jul 7, 2011 at 14:31
  • $\begingroup$ Who is this "younger brother" about whom you speak ?!?! (+1) $\endgroup$ Commented Jul 13, 2011 at 4:09
  • $\begingroup$ I had mixed things up when opening the square. Thanks for clarifying this, @Chandru. $\endgroup$
    – mathguy80
    Commented Jul 13, 2011 at 5:27
7
$\begingroup$

An alternative solution:

Substituting

$$\tan\left(\frac{A}{2}\right)=\frac{\sin(A)}{1+\cos(A)}$$

and multiplying through to clear the fractions quickly reduces the statement to one which is trivially true. Then working backwards through the simplifications you have made gives the proof.

Edit (providing requested clarification):

When you make the above substitution, you have

$$\frac{1+\sin A - \cos A}{1+\sin A + \cos A} = \frac{\sin A }{1+\cos A}$$

So

$$ (1+\sin A - \cos A)(1+\cos A) = \sin A (1+\sin A + \cos A)$$

And cancelling terms gives

$$1-\cos^2 A = \sin^2 A$$

So, using that as the starting point for your proof, "uncancelling" terms, and dividing gives the required identity.

$\endgroup$
3
  • $\begingroup$ That's good, and it supports mathguy's feeling that the half-angle identity for tangent would be useful. My solution is for those of us who have forgotten that half-angle identity. $\endgroup$ Commented Jul 7, 2011 at 13:38
  • $\begingroup$ I tried that but ended up going in circles, ending up with the other identity $\frac{1 - \cos A}{\sin A}$ Can you clarify what you mean by clearing fractions? $\endgroup$
    – mathguy80
    Commented Jul 7, 2011 at 13:40
  • $\begingroup$ Amazingly, a Canadian province I shall not name had for a long time the following final exam grading policy. In proving a trigonometric identity, a student could not "mix" the sides. Any computation step had to deal with one side or the other! $\endgroup$ Commented Jul 7, 2011 at 20:06
7
$\begingroup$

There is a theorem$^1$ that is worth knowing (I used it in this answer to this question) that states:

All direct trigonometric functions of $A$ ($2\alpha$) can be expressed rationally in terms of the tangent of $\frac{A}{2}$ ($\alpha$).

Combining $\sin A=2\sin \frac{A}{2}\cdot \cos \frac{A}{2}$ and $\cos ^{2} \frac{A}{2}+\sin ^{2}\frac{A}{2}=1$, we get (for $\cos \frac{A}{2}\neq 0$)

$$\sin A=\frac{2\sin \frac{A}{2}\cdot \cos \frac{A}{2}}{\cos ^{2}\frac{A}{2} +\sin ^{2}\frac{A}{2}}=\frac{2\tan \frac{A}{2}}{1+\tan ^{2}\frac{A}{2}}; \qquad (1)$$

and from $\cos A=\cos ^{2}\frac{A}{2}-\sin ^{2}\frac{A}{2}$, for $\cos \frac{A}{2}\neq 0$,

$$\cos A=\frac{\cos ^{2}\frac{A}{2}-\sin ^{2}\frac{A}{2}}{\cos ^{2}\frac{A}{2} +\sin ^{2}\frac{A}{2}}=\frac{1-\tan ^{2}\frac{A}{2}}{1+\tan ^{2}\frac{A}{2}}. \qquad (2)$$

To prove the identity multiply the LHS numerator and denominator by $1+\tan ^{2}\frac{A}{2}$ and use $(1)$ and $(2)$:

$$\begin{eqnarray*} \frac{1+\sin A-\cos A}{1+\sin A+\cos A} &=&\frac{1+2\tan \frac{A}{2}-1+\tan ^{2}\frac{A}{2}}{1+2\tan \frac{A}{2}+1+\tan ^{2}\frac{A}{2}} \\ &=&\frac{2\tan ^{2}\frac{A}{2}+2\tan \frac{A}{2}}{2+2\tan \frac{A}{2}}=\tan \frac{A}{2}. \end{eqnarray*}$$

--

$^1$ J. Calado, Compêndio de Trigonometria, 1967.

$\endgroup$
2
  • $\begingroup$ This question has got some excellent answers. This identity is new to me, I'm sure it will come in handy! Thank you! $\endgroup$
    – mathguy80
    Commented Jul 7, 2011 at 15:42
  • $\begingroup$ @mathguy80: You are welcome! $\endgroup$ Commented Jul 7, 2011 at 15:44
4
$\begingroup$

First write $A=2b$. Then make appropriate use of $\tan b=\sin b/\cos b$, $\sin2b=2\sin b\cos b$, and $\cos2b=2\cos^2b-1=1-2\sin^2b$.

$\endgroup$
1
  • $\begingroup$ Thanks. That's a neat substitution. It's a simple double angle identity from there. This also lead me backwards to, $\sin x = 2\sin \frac{x}{2} \cos \frac{x}{2}$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ $\endgroup$
    – mathguy80
    Commented Jul 7, 2011 at 13:38
3
$\begingroup$

The following is a method that in general I would not advocate as an approach to proving the usual trigonometric identities by hand. However, the idea has nice connections with other parts of mathematics, so it is worth mentioning.

There is a standard rational parametrization of (most of) the unit circle. Let $t=\tan(\theta/2)$. Then $$\cos \theta=\frac{1-t^2}{1+t^2} \qquad \text{and} \qquad \sin\theta=\frac{2t}{1+t^2}.$$ This parametrization gives a general method for integrating rational functions of $\sin\theta$ and $\cos\theta$. Parametrizations of the same general character are useful in number theory and elsewhere.

In our problem, the left-hand side is $$ \frac{1+\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}}{1+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}.$$ Bring top and bottom to the common denominator $1+t^2$. Things collapse, we get $t$. (As in the trigonometric proofs, we close our eyes to division by $0$.)

$\endgroup$
1
  • $\begingroup$ Another excellent answer. I had seen this substitution approach in some examples, but was unaware of what it was called. Thank you! $\endgroup$
    – mathguy80
    Commented Jul 7, 2011 at 15:44
1
$\begingroup$

This can also be done (mechanically) with complex numbers.

$\endgroup$
0
$\begingroup$

Using the identities$$1+\cos A=2 \cos ^{2} \frac{A}{2}, 1-\sin A=2 \sin ^{2} \frac{1}{2} , \sin A=2 \sin\frac{\theta}{2} \cos\frac{\theta}{2}, $$ We have $$ \begin{aligned} \frac{1+\sin A-\cos A}{1+\sin A+\cos A} &=\frac{2 \sin ^{2} \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos ^{2} \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}} \\ &=\frac{\sin \frac{A}{2}\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)}{\cot \frac{A}{2}\left(\cos \frac{A}{2}+\sin \frac{A}{2}\right)} \\ &=\tan \frac{A}{2} \end{aligned} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .