Question

Find the solution for the given trigonometric equation:
\[{\text{tan}}\theta + \tan 2\theta = \tan 3\theta \]

Answer 1

Hint: In this question, we will first convert the given function into sine and cosine function and then use trigonometric identities to the simplified expression. After this we will equate the different terms to zero to get the solution of each and finally combine them to get the solution for the given equation.

Complete step-by-step answer:
Given equation is:
\[{\text{tan}}\theta + \tan 2\theta = \tan 3\theta \]
Now, converting the tangent function in to sine and cosine function, we get:
$\dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\sin 2\theta }}{{\cos 2\theta }} = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$
Solving the left hand side, we get:
$\dfrac{{\sin \theta \times \cos 2\theta + \sin 2\theta \times \cos \theta }}{{\cos \theta \times \cos 2\theta }} = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$ (1)
We know that:
Sin (A+B) =$\sin A\cos B + \cos A\sin B$ .
So, using this identity equation 1 can be written as:
$\dfrac{{\sin (\theta + 2\theta )}}{{\cos \theta \cos 2\theta }} = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$ (2)
Now, using the identity,$\cos A\cos B = \dfrac{1}{2}\left( {\cos (A - B) + \cos (A + B} \right)$), we get:
$\dfrac{{\sin (\theta + 2\theta )}}{{\dfrac{1}{2}\left( {\cos ( - \theta ) + \cos 3\theta } \right)}} = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$
$ \Rightarrow \dfrac{{2\sin (3\theta )}}{{\left( {\cos ( - \theta ) + \cos 3\theta } \right)}} = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$
We know that: $\cos ( - \theta ) = \cos \theta $

Therefore, the above equation can written as
$\dfrac{{2\sin (3\theta )}}{{\left( {\cos \theta + \cos 3\theta } \right)}} = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$ (3)
Now, rearranging the equation 3, we get:
$\begin{gathered}
  \dfrac{{2\sin (3\theta )}}{{\left( {\cos (\theta ) + \cos 3\theta } \right)}} - \dfrac{{\sin 3\theta }}{{\cos 3\theta }} = 0 \\
    \\
\end{gathered} $
Taking the $\sin 3\theta $ term common, we get:
\[\sin 3\theta \left( {\dfrac{2}{{\left( {\cos (\theta ) + \cos 3\theta } \right)}} - \dfrac{1}{{\cos 3\theta }}} \right) = 0\]
Since, the products of two terms are zero. So the individual term will also be zero.
$\therefore \sin 3\theta = 0$ (4)
   and \[\left( {\dfrac{2}{{\left( {\cos (\theta ) + \cos 3\theta } \right)}} - \dfrac{1}{{\cos 3\theta }}} \right) = 0\] (5)
Solving equation 4, we get:
$3\theta = n\pi $
$ \Rightarrow \theta = \dfrac{{n\pi }}{3}$ (6)
Solving the equation 5, we get:
\[\dfrac{2}{{\left( {\cos (\theta ) + \cos 3\theta } \right)}} = \dfrac{1}{{\cos 3\theta }}\]
\[ \Rightarrow 2\cos 3\theta = \cos \theta + \cos 3\theta \]
\[ \Rightarrow \cos 3\theta = \cos \theta \]
\[ \Rightarrow \cos 3\theta - \cos \theta = 0\]
Using the difference to product identity of cosine function
$\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ , we get:
$2\sin 2\theta \sin \theta = 0$
$\therefore \sin 2\theta = 0$ (7)
Also
$\sin \theta = 0$ (8)
Solving equation 7, we get:
$2\theta = n\pi $
$ \Rightarrow \theta = \dfrac{{n\pi }}{2}$ (9)
We know that tangent function is not defined at $\theta = \dfrac{{n\pi }}{2}$
So, the solution given by equation 9 is not the solution for the given question.
Now, solving equation 8, we get
$\theta = n\pi $ (10)
Therefore, the solution for given equation are given by:
$\theta = \dfrac{{n\pi }}{3}$ and $\theta = n\pi $.

Note: In this type question, you can simplify the given equation by converting it into tangent function only and then use the trigonometric identity of tan(A+B)=$\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ to further simplify it and get the equation in product form as $\tan \theta \tan 2\theta (\tan \theta + \tan 2\theta ) = 0$ and then finally find the solution by equating each term to zero. You should never forget to recheck the solution.