Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $

Question

I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in \mathbb{Z} $)

Here is what I've done

$$\frac {\sin \theta}{\cos \theta} + \frac {\sin (2\theta)} {\cos (2\theta)} + \frac{\sin (3\theta)}{\cos (3\theta)}= \frac {\sin \theta}{\cos \theta} \frac {\sin (2\theta)} {\cos (2\theta)} \frac{\sin (3\theta)}{\cos (3\theta)}$$

$$ \frac {\sin \theta \cos (2\theta) \cos (3\theta) + \cos \theta \sin (2\theta) \cos (3\theta) + \cos \theta \cos (2\theta) \sin (3\theta) - \sin \theta \sin (2\theta) \sin (3\theta) }{\cos\theta \cos (2\theta) \cos (3\theta)} = 0 $$

$$\cos (2\theta) \{\sin\theta \cos (3\theta) + \cos \theta \sin (3\theta) \} + \sin (2\theta) \{\cos \theta \cos (3\theta) - \sin \theta \sin (3\theta) \} = 0 $$

$$\cos (2\theta) \sin(3\theta + \theta) +\sin(2\theta) \cos(3\theta + \theta) = 0 $$

$$ \cos (2\theta) \sin (4\theta) + \sin (2\theta) \cos (4\theta) = 0$$

$$ \sin (2\theta + 4\theta) = 0$$

$$\sin (6\theta) = 0 $$

$$ \theta = \frac {n\pi}{6}, n \in Z$$

I understand from this question that whatever mistake I am making is in the third step, where I remove $\cos \theta \cos (2\theta) \cos (3\theta) $ from the denominator. However, despite reading through the aforementioned post, I couldn't really get the intuition behind why this is wrong.

I'd like :

1. To understand the intuition behind why removing $\cos \theta \cos (2\theta) \cos (3\theta) $ is a mistake.
2. To know how to solve this question correctly
3. How do I avoid making these types of mistakes when solving trigonometric equations

Answer 1

In your simplified expression there is $\cos 3 \theta $ term in the denominator that goes to zero for solution you obtained. Should be checked before accepting or discarding it as a valid solution.

I cannot resist an elementary trig approach ..

If $ (A+B+C) = 2 \pi, $ then $ {(\tan A + \tan B + \tan C) = \tan A \tan B \tan C} $

If $ {(\tan A + \tan B + \tan C) = \tan A \tan B \tan C} $ then $(A+B+C) = 2 \pi $

is among possible solutions.

In the above if $ A= t, B=2t, C=3t$ then $t= \pm 2 \pi/6= \pm \pi/3,\pm 2\pi/3, ... $

By inspection $ t= 0 ,\,2 k\pi,$ plus co-terminals

Answer 2

You could have solved the problem using the tangent multiple angle formulae.

Using $t=\tan(\theta)$ $$\tan(2\theta)=\frac{2t}{1-t^2}\qquad \tan(3\theta)=\frac{3t-t^3}{1-3t^2}$$ This makes, after some minor simplifications

$$\tan \theta + \tan 2\theta + \tan 3\theta - \tan \theta \tan 2\theta \tan 3\theta=\frac{2 t \left(3-t^2\right)}{1-t^2}=0$$ and then the solutions assuming that ${1-3t^2}\neq 0$ and ${1-t^2}\neq 0$.

Answer 3

your intuition should be correct here's why

$\left(\frac{\sin\theta}{\cos\theta}+\frac{\sin2\theta}{\cos2\theta }+\frac{\sin3\theta}{\cos3\theta}=\frac{\sin\theta\sin2\theta\sin3\theta}{\cos\theta\cos2\theta\cos3\theta}\right)*\cos\theta\cos2\theta\cos3\theta$

$\Rightarrow \sin\theta\cos2\theta\cos3\theta+\cos\theta\sin2\theta\cos3\theta+\cos\theta\cos2\theta\sin3\theta= \sin\theta\sin2\theta\sin3\theta$

$\Rightarrow \cos\theta(\sin2\theta\cos3\theta+\cos2\theta\sin3\theta )+\sin\theta(\cos2\theta\cos3\theta-\sin2\theta\sin3\theta)=0 $

$\Rightarrow \cos\theta\sin(2\theta+3\theta)+\sin\theta\cos(2\theta+3\theta)=0 $

$\Rightarrow sin([2\theta+3\theta]+\theta)=0 $

$\Rightarrow sin(6\theta)=0 $

So, I'm uncertain that your answer is wrong...

Answer 4

We have $$\tan (A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1-(\tan A \tan B + \tan B \tan C + \tan C \tan A)}$$ and the given condition implies that $\tan (6\theta) = 0$. Thus $\theta = \frac{n\pi}{6}$, $n \in \mathbb{Z}$