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Question: Solve the given trigonometric equation: \(\tan 3\theta + \tan \theta = 2\tan 2\theta \)....

Solve the given trigonometric equation: tan⁡3θ+tan⁡θ=2tan⁡2θ\tan 3\theta + \tan \theta = 2\tan 2\theta .

Explanation

Solution

Before attempting this question, one should have prior knowledge about the trigonometric identities and also remember to use sin⁡(a+b)=sin⁡acos⁡b−sin⁡bcos⁡a\sin (a + b) = \sin a\cos b - \sin b\cos a, using this information can help you to approach the solution of the question.

Complete step-by-step solution:
According to the question, it is given that
tan⁡3θ+tan⁡θ=2tan⁡2θ\tan 3\theta + \tan \theta = 2\tan 2\theta
So, we can write 2tan⁡2θ=tan⁡2θ+tan⁡2θ2\tan 2\theta = \tan 2\theta + \tan 2\theta ............... (equation 1)
Now we can we can shuffle the equation by putting the value of 2tan⁡2θ2\tan 2\theta in equation 1, we get
⇒tan⁡3θ+tan⁡θ=2tan⁡2θ ⇒tan⁡3θ+tan⁡θ=tan⁡2θ+tan⁡2θ ⇒tan⁡3θ−tan⁡2θ=tan⁡2θ−tan⁡θ  \Rightarrow \tan 3\theta + \tan \theta = 2\tan 2\theta \\\ \Rightarrow \tan 3\theta + \tan \theta = \tan 2\theta + \tan 2\theta \\\ \Rightarrow \tan 3\theta - \tan 2\theta = \tan 2\theta - \tan \theta \\\
\Rightarrow $$$\dfrac{{\sin 3\theta }}{{\cos 3\theta }} - \dfrac{{\sin 2\theta }}{{\cos 2\theta }} = \dfrac{{\sin 2\theta }}{{\cos 2\theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}$ \Rightarrow \dfrac{{\sin 3\theta \cos 2\theta - \cos 3\theta \sin 2\theta }}{{\cos 3\theta \cos 2\theta }} = \dfrac{{\sin 2\theta \cos \theta - \cos 2\theta \sin \theta }}{{\cos 2\theta \cos \theta }}$$ As we know that by the trigonometric identities $$\sin (a + b) = \sin a\cos b - \sin b\cos a$$ therefore $$ \Rightarrow $$$\dfrac{{\sin \left( {3\theta - 2\theta } \right)}}{{\cos 3\theta \cos 2\theta }} = \dfrac{{\sin \left( {2\theta - \theta } \right)}}{{\cos 2\theta \cos \theta }}$ Cross multiplying both side we get $$ \Rightarrow \sin \theta \cos \theta = \sin \theta \cos 3\theta \Rightarrow $$sin⁡θcos⁡θ−sin⁡θcos⁡3θ=0\sin \theta \cos \theta - \sin \theta \cos 3\theta = 0
Or,sin⁡θ(cos⁡θ−cos⁡3θ)=0\sin \theta \left( {\cos \theta - \cos 3\theta } \right) = 0
We know that cos⁡a−cos⁡b=−2sin⁡(a+b2)sin⁡(a−b2)\cos a - \cos b = - 2\sin \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)
Therefore sin⁡θ⋅2sin⁡θsin⁡2θ=0\sin \theta \cdot 2\sin \theta \sin 2\theta = 0
So, sin⁡2θ=0{{\sin}^{2}} \theta = 0
Therefore θ=nπ\theta = n\pi
And also, sin⁡2θ=0\sin 2\theta = 0
Therefore 2θ=nπ2\theta = n\pi
⇒\Rightarrow θ=nπ2\theta = \dfrac{{n\pi }}{2}
But θ=nπ2\theta = \dfrac{{n\pi }}{2} will be rejected as when n is odd as it makes ∞=∞\infty = \infty
Therefore θ=nπ\theta = n\pi as it will form 0 = 0

Note: These trigonometric based problems as the trick to proving trigonometric identities are intuition, which can only be gained through experience. The more basic formulas you have memorized, the faster you will be. The following identities are essential to all your work with Trigo functions. Make a point of memorizing them.