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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: $$ a+b=\frac{\pi}{2} $$

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    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?2012-07-30
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    Should the denominator on the right be $1+\sin(\mathbf{2}a+2b)$?2012-07-30
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    What is the source of the problem? Why do you think it's true?2012-07-30
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    @Zander I've checked in Mathematica, this is true.2012-07-31
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    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 it2012-07-31
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    @daniel:fix $a$, the uniqueness of $b$ remain unknown2012-07-31
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    @daniel both are shown below...2012-07-31

2 Answers 2

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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):

$\hskip1.7in$enter image description here

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: $$ \begin{eqnarray} \frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos({\frac\pi2-a})} &=&\frac{1-\cos{(2a+\pi-2a)}}{1+\sin{(a+\pi-2a)}}\\ &=&\frac{1-\cos{(\pi)}}{1+\sin{(\pi-a)}}\\ \frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\sin a}&=&\frac{2}{1+\sin{a}}\\ \frac{2}{1+\sin{a}}&=&\frac{2}{1+\sin{a}}\\ \end{eqnarray} $$

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    @daniel ask WA, screenshot and upload (there is a small picture above the edit field)...2012-07-31
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    @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...2012-08-02
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    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.2012-08-02
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    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[\ $.2012-08-02
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    @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...2012-08-02
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    Thank you very much for your reply! I also used wolfram to solve it by turning $\tan{\frac{a}{2}},\tan{\frac{b}{2}}$ into variables, which also require much calculation. Your answer gave me another method to deal with this problem.2012-08-02
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    @Golbez You're welcome.2012-08-02
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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 $t=\tan{(0)}=0$.