Problem with goniometric function and triangles.

Question

I can't resolve this problem, above there is what I've done. Can you help me? Is there a better way to solve this problem?

A triangle $\triangle{ABC}$ is given, where the angle $\angle{BAC}=2{\alpha}$ has $\cos\angle{BAC}=\frac{7}{25}$ and the bisector $AD$ measures $2$.

1. Put $\angle{ADB} = x$ and prove that the area of the triangle $\triangle{ABC}$ can be expressed by the function $f(x)=\frac{48}{(16-9\cot^{2}(x))}$. Determine the triangle with the minimum area.

2. Write down the analytic expression of the function $g(x)=\left(\frac{96}{25}\right)\left(\frac{f(x)}{BC^{2}}\right)$ and draw the respective graphic.

Answer 1

Using the information given by the author,

$\mathsf{sin(\alpha)=\dfrac{3}{5}\,,\,\,cos(\alpha)=\dfrac{4}{5}\,,\,\,AB=\dfrac{2\,sin(x)}{sin(x+\alpha)}\,,\,\,AC=\dfrac{2\,sin(x)}{sin(x-\alpha)}\,\,\&}$

$\mathsf{sin(2\alpha)=\sqrt{1-cos^2(2\alpha)}=\sqrt{1-\dfrac{49}{625}}=\sqrt{\dfrac{625-49}{625}}=\dfrac{24}{25}}$

Now,

$\mathrm{Ar\left(\triangle ABC\right)=\dfrac{1}{2}\cdot{AB}\cdot{AC}\cdot sin(2\alpha)}$

$\mathrm{\implies Ar\left(\triangle ABC\right)=\dfrac{1}{2}\cdot\dfrac{2\,sin(x)}{sin(x+\alpha)}\cdot\dfrac{2\,sin(x)}{sin(x-\alpha)}\cdot sin(2\alpha)}$

$\mathrm{\implies Ar\left(\triangle ABC\right)=\dfrac{2\,sin^2(x)}{\left(sin(x)\,cos(\alpha)+cos(x)\,sin(\alpha)\right)\cdot \left(sin(x)\,cos(\alpha)-cos(x)\,sin(\alpha)\right)}\cdot \dfrac{24}{25}}$

$\mathrm{\implies Ar\left(\triangle ABC\right)=\dfrac{2\,sin^2(x)}{sin^2(x)\,cos^2(\alpha)-cos^2(x)\,sin^2(\alpha)}\cdot \dfrac{24}{25}}$

$\mathrm{\implies Ar\left(\triangle ABC\right)=\dfrac{48}{25\,cos^2(\alpha)-25\,sin^2(\alpha)\cdot cot^2(x)}}$

Put the values of $\mathrm{sin(\alpha)}$ and $\mathrm{cos(\alpha)}$

$\mathrm{\implies Ar\left(\triangle ABC\right)=f(x)=\dfrac{48}{16-9\,cot^2(x)}}$

Part(II):

Using laws of sine,

$\mathtt{\dfrac{BD}{sin(\alpha)}=\dfrac{AD}{sin(\pi-\alpha-x)}}$

$\implies \boxed{\mathtt{BD=\dfrac{2\,sin(\alpha)}{sin(x+\alpha)}}}$

And,

$\mathtt{\dfrac{DC}{sin(\alpha)}=\dfrac{AD}{sin(x-\alpha)}}$

$\implies \boxed{\mathtt{DC=\dfrac{2\,sin(\alpha)}{sin(x-\alpha)}}}$

Now,

$\mathsf{BC=BD+DC=\dfrac{2\,sin(\alpha)}{sin(x+\alpha)}+\dfrac{2\,sin(\alpha)}{sin(x-\alpha)}}$

$\mathsf{\implies\,BC=2\,sin(\alpha)\left\{\dfrac{sin(x-\alpha)+sin(x+\alpha)}{sin(x+\alpha)\cdot sin(x-\alpha)}\right\}}$

$\mathsf{\implies\,BC=\dfrac{2\,sin(x)\cdot2\,sin(\alpha)\,cos(\alpha)}{sin(x+\alpha)\cdot sin(x-\alpha)}}$

$\mathsf{\implies\,BC=\dfrac{2\,sin(x)\,sin(2\alpha)}{sin(x+\alpha)\cdot sin(x-\alpha)}}$

$\mathsf{\implies\,BC=\dfrac{1}{sin(x)}\cdot\color{green}{\dfrac{2\,sin^2(x)\,sin(2\alpha)}{sin(x+\alpha)\cdot sin(x-\alpha)}}}$

The highlighted expression is the area of the triangle, which is f(x)

So,

$\mathsf{\implies\,BC=cosec(x)\cdot\,f(x)}$

So,

$\mathsf{g(x)=\dfrac{96}{25}\cdot\dfrac{f(x)}{{(BC)}^{2}}}$

$\mathsf{\implies g(x)=\dfrac{96}{25}\cdot\dfrac{f(x)}{cosec^2(x)\cdot\,\left\{f(x)\right\}^2}}$

$\mathsf{\implies g(x)=\dfrac{96}{25}\cdot\dfrac{sin^2(x)}{f(x)}}$

Put the value of f(x),

$\mathsf{\implies g(x)=\dfrac{96}{25}\cdot\dfrac{sin^2(x)}{\dfrac{48}{16-9\,cot^2(x)}}}$

$\mathsf{\implies g(x)=\dfrac{96}{25}\cdot\dfrac{sin^2(x)\cdot\{16-9\,cot^2(x)\}}{48}}$

$\mathsf{\implies g(x)=\dfrac{96}{25}\cdot\dfrac{16\,sin^2(x)-9\,cos^2(x)}{48}}$

$\mathsf{\implies g(x)=\dfrac{2}{25}\cdot\left(16\,sin^2(x)-9\,cos^2(x)\right)}$

$\mathsf{\implies g(x)=\dfrac{32}{25}\,sin^2(x)-\dfrac{18}{25}\,cos^2(x)}$