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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?

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  • $\begingroup$ There is a mistake in your formula, the denominator should read $$3\cos^2A\color{red}{-}3.$$ $\endgroup$ – b00n heT Jul 24 '16 at 10:01
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$$k=\dfrac{\tan3A}{\tan A}=\dfrac{3-\tan^2A}{1-3\tan^2A}$$

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

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

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  • $\begingroup$ Wow. thsi one is really helpful. Thank you. $\endgroup$ – user355914 Jul 25 '16 at 10:43
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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 $$\frac{2k}{k-1} = \frac{2\sin 3A}{\sin 3A - \sin A + 4\sin^2 A} = \frac{2\sin 3A}{2\sin A} = \frac{\sin 3A}{\sin A}$$

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

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