Question

Prove the following trigonometric equation:
${{\cos }^{2}}x+{{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( x-\dfrac{\pi }{3} \right)=\dfrac{3}{2}$

Answer 1

Hint:We are going to solve the L.H.S of the given equation. We are going to use the identity of$\cos 2\theta =2{{\cos }^{2}}\theta -1$ in the L.H.S and after applying the value of ${{\cos }^{2}}x,{{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right),{{\cos }^{2}}\left( x-\dfrac{\pi }{3} \right)$ from the identity then the expression has terms which are in the form of $\cos C+\cos D$ so we will use that identity and then simplify.

Complete step-by-step answer:
The equation that we have to prove is:
${{\cos }^{2}}x+{{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( x-\dfrac{\pi }{3} \right)=\dfrac{3}{2}$
We are going to solve the L.H.S of the above equation and we get,
${{\cos }^{2}}x+{{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( x-\dfrac{\pi }{3} \right)$
If you can see carefully, the above expression has ${{\cos }^{2}}\theta$ kind of expressions so we can apply the double angle of cosine in the above problem.
$\begin{aligned} & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\\ & \Rightarrow {{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2} \\\ \end{aligned}$
Applying this double angle in ${{\cos }^{2}}x$ we get,
${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$
Applying this double angle in ${{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right)$ we get,
${{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right)=\dfrac{1+\cos \left( 2x+\dfrac{2\pi }{3} \right)}{2}$
Applying this double angle in ${{\cos }^{2}}\left( x-\dfrac{\pi }{3} \right)$ we get,
${{\cos }^{2}}\left( x-\dfrac{\pi }{3} \right)=\dfrac{1+\cos \left( 2x-\dfrac{2\pi }{3} \right)}{2}$
Substituting all these square of cosines value in the L.H.S of the given expression we get,
${{\cos }^{2}}x+{{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( x-\dfrac{\pi }{3} \right)$
$=\dfrac{3}{2}+\dfrac{1}{2}\left( \cos 2x+\cos \left( 2x+\dfrac{2\pi }{3} \right)+\cos \left( 2x-\dfrac{2\pi }{3} \right) \right)………..Eq.(1)$
Now, we are going to apply $\cos C+\cos D$ in the above equation we get,
$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
$\cos \left( 2x+\dfrac{2\pi }{3} \right)+\cos \left( 2x-\dfrac{2\pi }{3} \right)=2\cos 2x\cos \left( \dfrac{2\pi }{3} \right)$
Substituting the above value of cosines in eq. (1) we get,
$\dfrac{3}{2}+\dfrac{1}{2}\times 2\left( \cos 2x+2\cos 2x\cos \left( \dfrac{2\pi }{3} \right) \right)$
We know that, the value of $\cos \dfrac{2\pi }{3}$ is equal to $-\dfrac{1}{2}$. So, putting this value of cosine in the above expression we get,
$\begin{aligned} & \dfrac{3}{2}+\left( \cos 2x-2\left( \cos 2x \right)\dfrac{1}{2} \right) \\\ & =\dfrac{3}{2} \\\ \end{aligned}$
From the above simplification of L.H.S of the given equation, the value comes out to be $\dfrac{3}{2}$ which is equal to R.H.S of the given expression.
Hence, we have proved L.H.S = R.H.S of the given equation.

Note: You must be thinking that how do we know when to make a square of cosine to double angle as we have shown above.
${{\cos }^{2}}x+{{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( x-\dfrac{\pi }{3} \right)=\dfrac{3}{2}$
As you can see the R.H.S of the given equation, the value is $\dfrac{3}{2}$ and in the L.H.S. the angles are $\dfrac{\pi }{3}$ so if we double the angle of cosines in L.H.S then we get cosine of $\dfrac{2\pi }{3}$ as one of the trigonometric expressions then we are able to prove the given equation.