Question

If $\frac{tan3A}{tanA}=k$, then prove that $\frac{sin3A}{sinA}=\frac{2k}{k-1}$ and that k cannot lie between $\frac{1}{3}$ and 3.

Answer 1

$K-1=\frac{tan3A-tanA}{tanA}=\frac{sin2A}{cos3AsinA}=\frac{2cosA}{cos3A}$

$\frac{2K}{K-1}=\frac{2tan3A}{tanA}\cdot \frac{cos3A}{2cosA}=\frac{sin3A}{sinA}$

$\frac{\sin 3A}{\sin A}=\frac{3\sin A-4{\sin }^{3}A}{\sin A}=3-4{\sin }^{2}A$

Now we know

$0\le \sin A\le 1$

$\therefore -1\le 3-4{\sin }^{2}A\le 3$

$\therefore -1\le \frac{2K}{K-1}\le 3$

$-1\le \frac{2K}{K-1}\Rightarrow 1\ge \frac{2K}{1-K}\Rightarrow 1-K\ge 2K\Rightarrow 1\ge 3K\Rightarrow \frac{1}{3}\ge K$

$3\ge \frac{2K}{K-1}\Rightarrow 3K-3\ge 2K\Rightarrow K\ge 3$

$\Rightarrow K\le \frac{1}{3}$ and $K\ge 3$