A trigonometry equation problem

Let $a,b \in \left(0,\frac{\pi}{2}\right)$, satisfying $$\frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos{b}}=\frac{1-\cos{(2a+2b)}}{1+\sin{(a+2b)}}$$ Prove that: $$a+b=\frac{\pi}{2}$$

-
What have you tried? Did you try to add the fractions on the left-hand side, or use the trigonometric addition formulas on the right-hand side? – Jesse Madnick Jul 30 '12 at 17:28
Should the denominator on the right be $1+\sin(\mathbf{2}a+2b)$? – Théophile Jul 30 '12 at 18:09
What is the source of the problem? Why do you think it's true? – Zander Jul 30 '12 at 23:29
@Zander I've checked in Mathematica, this is true. – Norbert Jul 31 '12 at 7:19
It is true, I have used mathematica to solve it, take $\tan{a/2}$ and $\tan{b/2}$ as variables, and the result is true, but I need some easier way to solve it – Golbez Jul 31 '12 at 10:22

The following calculations, show the non-uniqueness of the statement to be proven and therefore that the proposition is not true, when we forget the restriction $a,b \in \left(0,\frac{\pi}{2}\right)$ (Thanks to Christian). Then $a+b=\frac\pi2$ is a sufficient, but not a necessary condition for $f(a,b) = 0$, because of the existence of other zeros. $f(a,b)$ can be splitted in at least $3$ factors and so $f(a,b)=0$ implies $\frac12\left(a+b-\frac\pi 2\right)=\pi n$ and/or $\frac12(a-\frac\pi2)=\pi m$, plus some more weird solutions, where a nice general expression is currently lacking (or might not exist), although some particular values can be specified. Here it is:

When you plot $$\begin{eqnarray} f(a,b)&=&(1+\cos(b))(1+\sin(a+2b))(1-\cos 2a)\\ &+&(1+\sin a)(1+\sin(a+2b))(1+\cos2a)\\ &-&(1+\cos(b))(1+\sin a)(1-\cos(2a+2b))=0 \end{eqnarray}$$ you'll clearly see the $a+b=\frac{\pi}{2}$-line, among other possible solutions (some of them are briefly discussed below in the EDIT):

$\hskip1.7in$

W|A helped to expand it to $$\sin(\frac a2-\frac\pi4) \color{red}{\sin(\frac a2+\frac b2-\frac\pi 4)} \biggr\{ \sin(a-\frac52 b)+3 \sin(a-\frac b2)+3 \sin(a+\frac b2)\\+6 \sin(a+\frac32 b)-\sin(3 a+\frac32 b)+2 \sin(a+\frac52 b)-2 \cos(2 a-\frac b2)\\ -\cos(2 a+\frac b2)-3 \cos(2 a+\frac32 b)+7 \cos(\frac b2)+2 \cos(\frac32 b)+\cos(\frac52 b) \biggr\}=0 \tag{*}$$ and then it is obvious, that $\frac12\left(a+b-\frac\pi 2\right)=\pi n$ solves this equation, as well as $\frac12(a-\pi/2)=\pi m$ does. Other solutions are visible, but not obvious.

EDIT Wolfram kindly provides some more particular roots of the $\{\cdots\}$-part of $(*)$ like $$a=2\pi c_1+\pi \; \;, b=-4\left(\pi c_2+\tan^{-1}(x_k)\right),$$ where

The converse of the statement is much easier to proof: Substitute $b=\pi/2-a$ to get: $$\begin{eqnarray} \frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos({\frac\pi2-a})} &=&\frac{1-\cos{(2a+\pi-2a)}}{1+\sin{(a+\pi-2a)}}\\ &=&\frac{1-\cos{(\pi)}}{1+\sin{(\pi-a)}}\\ \frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\sin a}&=&\frac{2}{1+\sin{a}}\\ \frac{2}{1+\sin{a}}&=&\frac{2}{1+\sin{a}}\\ \end{eqnarray}$$

-
@daniel ask WA, screenshot and upload (there is a small picture above the edit field)... – draks ... Jul 31 '12 at 19:39
@daniel thank you very much for caring so much. I took up your thoughts about $\frac12\left(a+b-\frac\pi 2\right)=\pi n$, $\frac12(a-\pi/2)=\pi m$ and updated the figure. I'm not sure how to relate this back to the original question. Any concrete ideas for that? Maybe it's the heavy use of Wolfram. Some people might not like it. I do, I love it. Thanks again... – draks ... Aug 2 '12 at 7:22
Maybe add at the beginning: "The OP asks for proof that ...etc. The following calculations show that the proposition is in general not true. a+b=pi/2 is a sufficient but not a necessary condition for f(a,b) = 0, because of the existence of other zeros. The implication is therefore of the form f(a,b)=0 implies A and/or B and/or C...in which A,B,C,... are zeros of the equation." I think this was a neat solution, because most people probably saw the problem of non-uniqueness, but showing it clearly was another matter. – daniel Aug 2 '12 at 9:44
The picture shows that the original statement is in fact true. Note that the variables $a$ and $b$ were from the beginning restricted to the interval $\ \bigl]0,{\pi\over2}\bigr[\$. – Christian Blatter Aug 2 '12 at 11:01
@ChristianBlatter, well spotted, my brain must have been still in the warm up phase, when I first read the question and I never got back that far. I edited it to take that into account... – draks ... Aug 2 '12 at 11:05

Let, without loss of generality, $a=\frac{\pi}{2}-(b-c)$,

then by using the fact that you can suck up $\frac{\pi}{2}$ (and thereby change $\sin$ to $\cos$) as well as $\pi$ (and thereby change a sign of the trigonometric function) you can get rid of all $\pi$'s and $a$'s:

$$\frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos{b}}=\frac{1-\cos{(2a+2b)}}{1+\sin{(a+2b)}}$$

$$\Longrightarrow$$

$$\frac{1+\cos{(2(b-c))}}{1+\cos{(b-c)}}+\frac{1-\cos{(2(b-c))}}{1+\cos{b}}=\frac{1+\cos{(2c)}}{1+\cos{(b+c)}}$$

Now it is acutally clear that $a+b=\frac{\pi}{2}$ is a solution, because if you plug in $c=0$, all three denomiators become $1+\cos{(b)}$ and the equation reads $2=2$.

If you want to go futher you can use $$\cos ( b \pm c ) = \cos (b) \cos (c) \mp \sin (b) \sin (c)$$ to isolate the two trigonometric functions of $c$, substitue $t=\tan{\frac{c}{2}}$ to get polynomial expression in $t$, put everything on one equal denominator and solve the thing. This will amount to $t=\tan{(0)}=0$.

-