Question

Solve the equation \[\cos \theta .cos2\theta .\cos 3\theta =\dfrac{1}{4}\].

Answer 1

Hint: We will apply a trigonometric formula given by \[2\operatorname{cosAcosB}=\cos \left( A+B \right)+\cos \left( A-B \right)\] to simplify the terms in the given equation. Then we will again use other trigonometric formulas like \[cos2\theta =2{{\cos }^{2}}\theta -1\] to rearrange and simplify the terms further.

Complete step-by-step answer:

We will first consider the expression \[\cos \theta .cos2\theta .\cos 3\theta =\dfrac{1}{4}.....(1)\]

We will consider the left hand side of the equation (1).

\[\Rightarrow \cos \theta .cos2\theta .\cos 3\theta \]

By multiplying and dividing it by 2, we get,

\[\Rightarrow \dfrac{1}{2}\times 2\cos \theta .cos2\theta .\cos 3\theta =\dfrac{2}{2}\left( cos\theta cos3\theta \right)\cos 2\theta \]

By applying the formula \[2\operatorname{cosAcosB}=\cos \left( A+B \right)+\cos \left( A-B \right)\] to the expression \[2\cos \theta \cos 3\theta =\cos \left( \theta +3\theta \right)\cos \left( \theta -3\theta \right)\].

Therefore, we have,

\[\begin{align}

  & \cos \theta .\cos 2\theta .\cos 3\theta =\dfrac{\left[ \cos \left( \theta +3\theta \right)\cos \left( \theta -3\theta \right) \right]}{2}\cos 2\theta \\

 & \cos \theta .\cos 2\theta .\cos 3\theta =\left[ \cos 4\theta +\cos (-2\theta ) \right]\dfrac{\cos 2\theta }{2} \\

\end{align}\]

Since we know that \[-2\theta \] in \[\cos \left( -2\theta \right)=\cos 2\theta \]. Thus we get,
\[\cos \theta .\cos 2\theta .\cos 3\theta =\left[ \cos 4\theta +\cos 2\theta \right]\dfrac{\cos 2\theta }{2}\]

Now we will substitute the value of \[\cos \theta .\cos 2\theta .\cos 3\theta \] and write it in

equation (1). Therefore, we have,

\[\begin{align}

  & \left[ \cos 4\theta +cos2\theta \right]\dfrac{\cos 2\theta }{2}=\dfrac{1}{4} \\

 & \left[ \cos 4\theta +\cos 2\theta \right]\cos 2\theta =\dfrac{1}{2} \\

 & 2\left[ \cos 4\theta +\cos 2\theta \right]\cos \left( 2\theta \right)=\dfrac{1}{2}\times 2 \\

 & 2\left[ \cos 4\theta +\cos 2\theta \right]\cos 2\theta =1 \\

\end{align}\]

\[\begin{align}

  & 2\cos 4\theta \cos 2\theta +2{{\cos }^{2}}(2\theta )=1 \\

 & 2\cos 4\theta \cos 2\theta +2{{\cos }^{2}}(2\theta )-1=0 \\

\end{align}\]

Now we will apply the formula of \[{{\cos }^{2}}(2\theta )\].

We know that \[cos2\theta =2{{\cos }^{2}}\theta -1\]. So if we have \[\cos 4\theta =2{{\cos

}^{2}}\left( 2\theta \right)-1\].

Therefore, we have,

\[\begin{align}

  & 2\cos 4\theta \cos 2\theta +\left[ 2{{\cos }^{2}}\left( 2\theta \right)-1 \right]=0 \\

 & 2cos4\theta \cos 2\theta +\cos 4\theta =0 \\

 & \cos 4\theta \left[ 2\cos 2\theta +1 \right]=0 \\

\end{align}\]

Therefore, we have \[\cos 4\theta =0\] or \[\left[ 2\cos 2\theta +1 \right]=0\].

Since we know that \[0=\cos \dfrac{\pi }{2}\], therefore, we have that \[\cos 4\theta =\cos \dfrac{\pi }{2}\].

\[4\theta =\dfrac{\pi }{2}\] can’t be directly put here. The value of \[\cos \dfrac{\pi }{2}\] is positive in the first quadrant and negative in the second quadrant. So \[\cos 4\theta =\cos \dfrac{\pi }{2}\].

\[\Rightarrow 4\theta =\dfrac{\pi }{2}\] which is now correct.

 \[\begin{align}

  & \Rightarrow 4\theta =\dfrac{\pi }{2}+2n\pi \\

 & \theta =\dfrac{\pi }{8}+\dfrac{2n\pi }{4} \\

\end{align}\]

\[\theta =\dfrac{\pi }{8}+\dfrac{n\pi }{4}\], where \[n=\pm 1,\pm 2,...\]

Now we will consider the expression,

\[\begin{align}

  & 2\cos 2\theta +1=0 \\

 & 2\cos 2\theta =-1 \\

 & \cos 2\theta =\dfrac{-1}{2} \\

\end{align}\]

As we know that the value of \[\dfrac{1}{2}=\cos \dfrac{\pi }{3}\].

Therefore, we have that,

\[\cos 2\theta =-\cos \dfrac{\pi }{3}\]

We know that the value of cos is negative in the second and the third quadrant. Therefore,

we consider the second quadrant. In this quadrant,

\[\begin{align}

  & \cos \dfrac{\pi }{3}=\cos \left( \pi -\dfrac{\pi }{3} \right) \\

 & \cos 2\theta =\cos \left( \pi -\dfrac{\pi }{3} \right) \\

 & \cos 2\theta =\cos \left( \dfrac{2\pi }{3} \right) \\

\end{align}\]

\[2\theta =\dfrac{2\pi }{3}+2n\pi \], where \[n=\pm 0,\pm 1,\pm 2,...\]

or \[\theta =\dfrac{2\pi }{2.3}+\dfrac{2n\pi }{2}\]

\[\theta =\dfrac{\pi }{3}+n\pi \], where \[n=\pm 0,\pm 1,\pm 2,...\]

Also, we will consider the third quadrant in which,

\[\begin{align}

  & \cos \dfrac{\pi }{3}=\cos \left( \pi +\dfrac{\pi }{3} \right) \\

 & \cos 2\theta =\cos \left( \pi +\dfrac{\pi }{3} \right) \\

 & 2\theta =\dfrac{4\pi }{3} \\

\end{align}\]

or \[2\theta =\dfrac{4\pi }{3}+2n\pi \], where \[n=\pm 0,\pm 1,\pm 2,...\]

\[\theta =\dfrac{4\pi }{3.2}+\dfrac{2n\pi }{2}\], where \[n=\pm 0,\pm 1,\pm 2,...\]

\[\theta =\dfrac{2\pi }{3}+n\pi \], where \[n=\pm 0,\pm 1,\pm 2,...\]

Hence the value of \[\theta =\dfrac{\pi }{8},\dfrac{n\pi }{4},\dfrac{\pi }{3}+n\pi ,\dfrac{2\pi

}{3}+n\pi \].



Note: While solving this question we come across one term equal to \[-\cos \left( \dfrac{\pi }{3} \right)\]. Do not change this value to \[-\cos \dfrac{\pi }{3}=\cos \left( \dfrac{-\pi }{3} \right)\] which results into positive \[\cos \dfrac{\pi }{3}\]. This will hinder the actual value of the equation for \[\theta \]. By applying the formula we can have a new expression by \[2\cos A\cos B=\cos \left( A+B \right)\cos \left( A-B \right)\]. Since in the question, it is applied to \[2\cos \theta \cos 3\theta \], it’s because of the ignorance of the complex terms created by \[2\cos \theta \cos 2\theta \].