A trigonometry equation problem

Question

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

Answer 1

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

• $x_{0,1}=1\pm\sqrt{2}\;$ (roots of $x^2-2x-1\;$) or
• $x_{2,3,4,5}$, one of $4$ real roots of $x^8-8 x^7-20 x^6-8 x^5+22 x^4+8x^3-20 x^2+8 x+1$.

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

Answer 2

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 $t=\tan{(0)}=0$.