# Deduce the relation from the given trigonometric relation

If $$\frac{\tan3A}{\tan A}=k$$ Then prove that $$\frac{\sin3A}{\sin A} = \frac{2k}{k-1}$$

I tried this, $$\tan3A = \frac{3\tan A-\tan^3 A}{1-3\tan^2 A}$$ then divided by $\tan A$ on both sides and finally got $$k= \frac{4\cos^2 A-1}{4\cos^2 A + 3}$$

but I cannot do further. Can you explain, please?

• There is a mistake in your formula, the denominator should read $$3\cos^2A\color{red}{-}3.$$ – b00n heT Jul 24 '16 at 10:01

$$k=\dfrac{\tan3A}{\tan A}=\dfrac{3-\tan^2A}{1-3\tan^2A}$$

$$\implies\dfrac k{k-1}=\dfrac{3-\tan^2A}{3-\tan^2A-(1-3\tan^2A)}$$

Multiplying the numerator & the denominator by $\cos^2A,$

$$\dfrac k{k-1}=\dfrac{3\cos^2A-\sin^2A}2=\dfrac{3(1-\sin^2A)-\sin^2A}2=?$$

Now, $$\dfrac{\sin3A}{\sin A}=\dfrac{3\sin A-4\sin^3A}{\sin A}=?$$

• Wow. thsi one is really helpful. Thank you. – user355914 Jul 25 '16 at 10:43

We have \begin{align*}\frac{2k}{k-1} = \frac{2\frac{\tan 3A}{\tan A}}{\frac{\tan 3A}{\tan A} - 1} &= \frac{2\tan 3A}{\tan 3A - \tan A} \\ & = \frac{2\frac{\sin 3A}{\cos 3A}}{\frac{\sin 3A}{\cos 3A} - \frac{\sin A}{\cos A}} \\ & = \frac{2\sin 3A}{\sin 3A - \sin A \cdot \frac{\cos 3A}{\cos A}}\end{align*}

But we know that $\cos 3A = 4\cos^3 A - 3\cos A \Rightarrow \frac{\cos 3A}{\cos A} = 4\cos^2 A - 3 = 1 - 4\sin^2 A$, giving us $$\frac{2k}{k-1} = \frac{2\sin 3A}{\sin 3A - \sin A + 4\sin^2 A} = \frac{2\sin 3A}{2\sin A} = \frac{\sin 3A}{\sin A}$$

• Yes, I've understood. thanks sir. – user355914 Jul 24 '16 at 16:15