If $$m \tan(\theta - \pi/6) = n \tan(\theta + 2\pi/3)$$
then find $\cos 2\theta$ in terms of $m$ and $n$.
What is the correct method to solve this question?
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Jamil Ahmed
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1Well, what have you tried? Also, if do remember that Mathematics is not a pursuit for an easy method, but one that gives the most truest of outcomes. – Jun 17 '16 at 10:43
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2Possible duplicate of Prove that $m\tan (\theta-30°)=n\tan (\theta+120°)$ – Martin Sleziak Jun 17 '16 at 14:17
2 Answers
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$$\dfrac{\sin\left(\theta-\dfrac\pi6\right)\cos\left(\theta+\dfrac{2\pi}3\right)}{\cos\left(\theta-\dfrac\pi6\right)\sin\left(\theta+\dfrac{2\pi}3\right)}=\dfrac nm$$
Using Componendo and Dividendo and $\sin(A\pm B)$ formulae,
$$\dfrac{m-n}{m+n}=\dfrac{\sin\left(\dfrac{2\pi}3+\dfrac\pi6\right)}{\sin\left(2\theta+\dfrac{2\pi}3-\dfrac\pi6\right)}=\dfrac{\dfrac12}{\cos2\theta}$$ as $\sin\left(\dfrac{5\pi}6\right)=\sin\left(\pi-\dfrac\pi6\right)=\sin\dfrac\pi6=?$
and $\sin\left(A+\dfrac\pi2\right)=\cos A$
lab bhattacharjee
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+1 Looks good to me! Certainly gives the right answer for $m/n=1,2,3$. – almagest Jun 17 '16 at 11:26
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@almagest, Thanks for confirmation :) Not the toughest solution, right? – lab bhattacharjee Jun 17 '16 at 11:35
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$$m\left( \frac{\tan \theta -\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}\tan \theta } \right)=n\left( \frac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\tan \theta } \right)\,\,\,\,\Rightarrow \,\,\,{{\tan }^{2}}\theta =\frac{m-3n}{3m-n}$$ Now apply this identity $$\cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }$$
Behrouz Maleki
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There must be a mistake somewhere here. Put $m=3n$. Then the original equation has solution $\theta=n\pi$ or $\tan\theta=0$. But you get $\tan\theta=0.75$. – almagest Jun 17 '16 at 11:24
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