The problem is:
If one side and the opposite angle of a triangle are fixed, and the other two sides are variable, use the law of cosines to show that the area is a maximum when the triangle is isosceles.
I know that, if I let $c$ denote the fixed side and $\theta$ the angle opposite to $c$, then area $A=1/2 ab \sin{\theta}$, where $a$ and $b$ are the variable sides. I've tried substituting $b$ by the expression for it determined by the law of cosines, but that didn't seem to get me anywhere.
On
hint
Do not forget the useful equality
$$\frac {a}{\sin (\alpha)}=\frac {b}{\sin (\beta)}=\frac {c}{\sin (\gamma)} $$
On
Fix the side of length $c$. Let $C$ be the vertex of the triangle opposite to $c$. The set of points that can be chosen for $C$, forming a fixed angle $\theta$, forms a circle, such that $c$ is a chord of that circle. (Actually it forms two circles, one on each side, but we can just take one WLOG.)
The area of the triangle is $\frac12 c h$. So the question is what point $C$ on this circle maximizes the height $h$ perpendicular to $c$. Somewhat clearly, that would be the point $C$ such that the line tangent to the circle at $C$ is parallel to $c$. It follows from this that the triangle is isosceles.
Alternate approach: If you want to make your idea work, you would be trying to maximize $$ \left(\frac12 \sin \theta \right) ab $$ subject to the constraint $a^2 + b^2 - 2ab \cos \theta = c$, for a fixed $c$ and $\theta$. This is a Lagrange multipliers problem, with two variables $a$ and $b$.
On
I you use the extended version of the sine rule, with $a$ the side, $A$ the opposite angle and $R$ the circumradius, you find that $2R=\frac a{\sin A}$, so that the third vertex of the triangle lies on a fixed circle. The centre of the circle lies on the perpendicular bisector of the side you have been given (the circle passes through the two vertices you already have) and the greatest possible area is when the vertex also lies on the bisector of this side (greatest height on fixed base).
On
If $AB$ is fixed and $\widehat{ACB}$ is fixed, the circumradius of $ABC$ is fixed as well.
If $C$ travels on the (major, or minor) $AB$-arc of a fixed circle $\Gamma$, the distance of $C$ from $AB$, hence the area of $ABC$, is clearly maximized when $C$ lies on the perpendicular bisector of $AB$, hence when $ABC$ is isosceles with respect to the base $AB$.
Let $a$ be a fixed side and $\alpha$ be a fixed angle of $\Delta ABC$.
Thus, in the standard notation we obtain: $$S_{\Delta ABC}=\frac{1}{2}bc\sin\alpha=2R^2\sin\alpha\sin\beta\sin\gamma=$$ $$=R^2\sin\alpha(\cos(\beta-\gamma)-\cos(\beta+\gamma))\leq R^2\sin\alpha(1+\cos\alpha).$$ The equality occurs for $\cos(\beta-\gamma)=1$, which gives $\beta=\gamma$ and we are done!