Solving Triangles - Part 5

For any $x \in [0, \frac { \pi }{ 2 } ]$ and the function $f(x)=\frac { \sin {2x} }{ \sin {x}+\cos {x} }$ , the following is true:

$0\le f(x)\le \frac { \sqrt { 2 } }{ 2 }$

Let's prove that: since $x \in [0, \frac { \pi }{ 2 } ]$ , then $\sin { 2x }$ , $\sin { x }$ and $\cos { x }$ are positive. $f(x)\ge 0$ proved. Let's find the maximum through derivative:

$f'(x)=\frac { 2\cos { 2x } \left( \sin { x } +\cos { x } \right) -\sin { 2x } \left( \cos { x } -\sin { x } \right) }{ { \left( \sin { x } +\cos { x } \right) }^{ 2 } }$

To find the maximum we need to solve this equation:

$f'(x)=0\\ 2\cos { 2x } \left( \sin { x } +\cos { x } \right) -\sin { 2x } \left( \cos { x } -\sin { x } \right) =0\\ 2\cos { 2x } \left( \sin { x } +\cos { x } \right) =\sin { 2x } \left( \cos { x } -\sin { x } \right) \\ 2\left( { \cos }^{ 2 }{ x }-{ \sin }^{ 2 }{ x } \right) \left( \sin { x } +\cos { x } \right) =2\sin { x } \cos { x } \left( \cos { x } -\sin { x } \right) \\ { \cos }^{ 2 }{ x }\sin { x } +{ \cos }^{ 3 }{ x }-{ \sin }^{ 3 }{ x }-{ \sin }^{ 2 }{ x }\cos { x } =2\sin { x } { \cos }^{ 2 }{ x }-2\cos { x } { \sin }^{ 2 }{ x }\\ { \cos }^{ 3 }{ x }+{ \sin }^{ 2 }{ x }\cos { x } ={ \cos }^{ 2 }{ x }\sin { x } +{ \sin }^{ 3 }{ x }\\ \cos { x } \left( { \cos }^{ 2 }{ x }+{ \sin }^{ 2 }{ x } \right) =\left( { \cos }^{ 2 }{ x }+{ \sin }^{ 2 }{ x } \right) \sin { x } \\ \cos { x } =\sin { x } \\ x=\frac { \pi }{ 4 } ,\quad x \in [0, \frac { \pi }{ 2 } ]$

Then:

$\max { f(x) } =f(\frac { \pi }{ 4 } )=\frac { \sqrt { 2 } }{ 2 }$

That means if we prove that $0\le \angle B\le \frac { \pi }{ 2 }$ the problem is solved.

Since the sinuses of angles form successive terms of a geometric progression, we get:

$\begin{cases} \sin { A } =\frac { 1 }{ k } \cdot \sin { B } \\ \sin { C } =k\cdot \sin { B } \end{cases}$ , where $k$ is the common ratio.

Let's assume that the radius of the circumcircle equals 1. Then by sinus theorem:

$\frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } } =2R\\ \frac { a }{ \frac { 1 }{ k } \cdot \sin { B } } =\frac { b }{ \sin { B } } =\frac { c }{ k\cdot \sin { B } } =2$

And we have:

$\begin{cases} a=\frac { 2 }{ k } \cdot \sin { B } \\ b=2\cdot \sin { B } \\ c=2k\cdot \sin { B } \end{cases}$

Now we can use cosinus theorem:

${ b }^{ 2 }={ a }^{ 2 }+{ c }^{ 2 }-2ac\cos { B } \\ 4\sin ^{ 2 }{ B } =\frac { 4 }{ { k }^{ 2 } } \sin ^{ 2 }{ B } +4{ k }^{ 2 }\sin ^{ 2 }{ B } -8\sin ^{ 2 }{ B } \cos { B } \\ 1=\frac { 1 }{ { k }^{ 2 } } +{ k }^{ 2 }-2\cos { B } \\ \cos { B } =\frac { 1 }{ 2{ k }^{ 2 } } +\frac { { k }^{ 2 } }{ 2 } -\frac { 1 }{ 2 } =\frac { 1+{ k }^{ 4 }-{ k }^{ 2 } }{ 2{ k }^{ 2 } } \\ \cos { B } =\frac { { ({ k }^{ 2 }-\frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } }{ 2{ k }^{ 2 } } >0$

Thus, $\cos { B } >0$ and $\angle B$ is in triangle which means $\angle B$ is acute.

Examples:

$AC=1, f(B)=\frac { \sqrt { 2 } }{ 2 } ,\quad \angle B=\frac { \pi }{ 4 } ,\quad k=\sqrt { \frac { 1 }{ 2 } (1+\sqrt { 2 } +\sqrt { 2\sqrt { 2 } -1 } ) }$

$AC=1, f(x)\rightarrow +0,\quad \angle B\rightarrow +0,\quad k\rightarrow \frac { 1+\sqrt { 5 } }{ 2 }$