Question

We have a trigonometric expression $\cos \theta +\cos 7\theta +\cos 3\theta +\cos 5\theta =0$ Find the value of $\theta$.

Answer 1

This question can be solved by using the transformation formula and then getting all the possible solutions for the equations obtained.
$\cos c+\cos d=2\cos \left( \dfrac{c+d}{2} \right)\cos \left( \dfrac{c-d}{2} \right)$

Complete step-by-step solution:
The transformation formula that we should use is:
$\cos c+\cos d=2\cos \left( \dfrac{c+d}{2} \right)\cos \left( \dfrac{c-d}{2} \right)..........\left( 1 \right)$
Now, let us split the given equation into two parts and then apply the transformation formula from equation (1) to each part.
Let us consider the first part and simplify it:
$\Rightarrow \cos \theta +\cos 7\theta$
By comparing this with the transformation formula (1) we get,
$c=\theta ,d=7\theta$
Now, by substituting these values in the transformation formula (1) we get,
$\Rightarrow 2\cos \left( \dfrac{\theta +7\theta }{2} \right)\cos \left( \dfrac{\theta -7\theta }{2} \right)$
$\Rightarrow 2\cos \left( \dfrac{8\theta }{2} \right)\cos \left( \dfrac{-6\theta }{2} \right)$
By simplifying the above step we can get,
$\Rightarrow 2\cos 4\theta \cos \left( -3\theta \right)$
As we already know that from the properties of cosine :
$\Rightarrow 2\cos 4\theta \cos \left( 3\theta \right)\text{ }\left[ \because \cos \left( -\theta \right)=\cos \theta \right]$………… (2)
Now, let us consider the second part and solve it.
$\Rightarrow \cos 3\theta +\cos 5\theta$
By comparing this with the transformation formula (1) we get,
$c=3\theta ,d=5\theta$
Now, by substituting these values in the transformation formula (1):

& \Rightarrow 2\cos \left( \dfrac{3\theta +5\theta }{2} \right)\cos \left( \dfrac{3\theta -5\theta }{2} \right) \\\ & \Rightarrow 2\cos \left( \dfrac{8\theta }{2} \right)\cos \left( \dfrac{-2\theta }{2} \right) \\\ \end{aligned}$$ By simplifying the above step we can get, $$\Rightarrow 2\cos 4\theta \cos \left( -\theta \right)$$ As we already know that from the properties of cosine: $$\Rightarrow 2\cos 4\theta \cos \left( \theta \right)\text{ }\left[ \because \cos \left( -\theta \right)=\cos \theta \right]$$…………… (3) Let us now add the two equations (2) and (3) that are obtained by simplifying the two parts using the transformation formula. $$\Rightarrow \cos \theta +\cos 7\theta +\cos 3\theta +\cos 5\theta =0$$ We can rewrite it using (2) and (3) that we got by applying the transformation formula. $$\Rightarrow 2\cos 4\theta \cos \left( 3\theta \right)\text{+}2\cos 4\theta \cos \left( \theta \right)\text{=0}$$ Now, we can take $$\cos 4\theta $$common out of both the addition terms and rewrite it as: $$\Rightarrow 2\cos 4\theta \left( \cos 3\theta +\cos \theta \right)\text{=0 }$$ Here, we need to split it into two parts to get the solution because there are two possibilities: $$\cos 4\theta =0\text{ or }\cos 3\theta +\cos \theta =0$$ First let us write general solution for the first one: $$\Rightarrow \cos 4\theta =0$$ From, the results of trigonometric equations we know that, $$\cos \theta =0\Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{2}...........\left( 4 \right)$$ So, from the result obtained in equation (4) we get, $$\begin{aligned} & \Rightarrow 4\theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\\ & \therefore \theta =\left( 2n+1 \right)\dfrac{\pi }{8} \\\ \end{aligned}$$ $$\left[ \because n\ge 0 \right]$$ $$\therefore \theta =\dfrac{\pi }{8},\dfrac{3\pi }{8},\dfrac{5\pi }{8}etc$$ Now, let us write general solution for second one: $$\Rightarrow \cos 3\theta +\cos \theta =0$$ By applying the transformation equation to this again we get, $$\begin{aligned} & \Rightarrow \cos 3\theta +\cos \theta =0 \\\ & \Rightarrow 2\cos \left( \dfrac{3\theta +\theta }{2} \right)\cos \left( \dfrac{3\theta -\theta }{2} \right)=0 \\\ & \Rightarrow 2\cos \left( \dfrac{4\theta }{2} \right)\cos \left( \dfrac{2\theta }{2} \right)=0 \\\ & \Rightarrow 2\cos 2\theta \cos \theta =0 \\\ \end{aligned}$$ This again implies that: $$\cos 2\theta =0\text{ or }\cos \theta =0$$ From equation (4) we can write, $$\begin{aligned} & \Rightarrow 2\theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\\ & \therefore \theta =\left( 2n+1 \right)\dfrac{\pi }{4} \\\ & \text{ }\therefore \theta =\frac{\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4}etc \\\ \end{aligned}$$ $$\left[ \because n\ge 0 \right]$$ And also from other one: $$\begin{aligned} & \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\\ & \therefore \theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\\ & \text{ }\therefore \theta =\frac{\pi }{2},\dfrac{3\pi }{2},\dfrac{5\pi }{2}etc \\\ \end{aligned}$$ $$\left[ \because n\ge 0 \right]$$ Hence, value of $$\theta =\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{5\pi }{2},\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4},\dfrac{\pi }{8},\dfrac{3\pi }{8},\dfrac{5\pi }{8}etc$$ **Note:** Instead of using the transformation formula we can also do it by using the trigonometric ratios of compound angles but it may take little more time whereas the conclusion will be the same. $$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$$ While equating to zero when the terms are in multiplication it is necessary to consider that either of the terms or both can be zero while neglecting any of these leads to wrong results. Writing the general solution helps in finding the solutions easily so that we can substitute the value of n and can get the values.