Question

In a $\Delta ABC$, prove that ${{\cos }^{2}}A+{{\cos }^{2}}\left( A+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( A-\dfrac{\pi }{3} \right)=\dfrac{3}{2}$.

Answer 1

Hint: For solving this question first we will apply the trigonometric formulas for sum and difference of two angles and some standard trigonometric ratios like $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ and $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$. After that, we will apply the whole square formula to simplify the term on the left-hand side and prove it equal to the term on the right-hand side.

Complete step-by-step answer:

Given:

For a $\Delta ABC$ we have to prove the following equation:

${{\cos }^{2}}A+{{\cos }^{2}}\left( A+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( A-\dfrac{\pi }{3} \right)=\dfrac{3}{2}$

Now, before we proceed we should know the following seven formulas:

$\begin{align}

  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1................................\left( 1 \right) \\

 & \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B..................\left( 2 \right) \\

 & \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B...................\left( 3 \right) \\

 & \cos \dfrac{\pi }{3}=\dfrac{1}{2}..............................................................\left( 4 \right) \\

 & \sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}............................................................\left( 5 \right)
\\

 & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab..........................................\left( 6 \right) \\

 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab...........................................\left( 7 \right) \\

\end{align}$

Now, we will be using the above seven formulas to simplify the term on the left-hand side to prove the desired result.

Now, L.H.S is equal to ${{\cos }^{2}}A+{{\cos }^{2}}\left( A+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( A-\dfrac{\pi }{3} \right)$ so, using the formula from the equation (2) and equation (3). Then,

$\begin{align}

  & {{\cos }^{2}}A+{{\cos }^{2}}\left( A+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( A-\dfrac{\pi }{3} \right) \\

 & \Rightarrow {{\cos }^{2}}A+{{\left( \cos A\cos \dfrac{\pi }{3}-\sin A\sin \dfrac{\pi }{3} \right)}^{2}}+{{\left( \cos A\cos \dfrac{\pi }{3}+\sin A\sin \dfrac{\pi }{3} \right)}^{2}} \\

\end{align}$

Now, substituting the value of $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ form equation (4) and $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$ from equation (5) in the above equation. Then,

$\begin{align}

  & {{\cos }^{2}}A+{{\left( \cos A\cos \dfrac{\pi }{3}-\sin A\sin \dfrac{\pi }{3} \right)}^{2}}+{{\left( \cos A\cos \dfrac{\pi }{3}+\sin A\sin \dfrac{\pi }{3} \right)}^{2}} \\

 & \Rightarrow {{\cos }^{2}}A+{{\left( \dfrac{\cos A}{2}-\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}}+{{\left( \dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}} \\

\end{align}$

Now, using the formula from the equation (6) and equation (7) in the above equation. Then,

$\begin{align}

  & {{\cos }^{2}}A+{{\left( \dfrac{\cos A}{2}-\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}}+{{\left( \dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}} \\

 & \Rightarrow {{\cos }^{2}}A+\dfrac{{{\cos }^{2}}A}{4}+\dfrac{3{{\sin }^{2}}A}{4}-2\times \dfrac{\cos A}{2}\times \dfrac{\sqrt{3}\sin A}{2}+\dfrac{{{\cos }^{2}}A}{4}+\dfrac{3{{\sin }^{2}}A}{4}+2\times \dfrac{\cos A}{2}\times \dfrac{\sqrt{3}\sin A}{2} \\

 & \Rightarrow \dfrac{3{{\cos }^{2}}A}{2}+\dfrac{3{{\sin }^{2}}A}{2} \\

 & \Rightarrow \dfrac{3}{2}\left( {{\cos }^{2}}A+{{\sin }^{2}}A \right) \\

\end{align}$

Now, substituting the value of ${{\cos }^{2}}A+{{\sin }^{2}}A=1$ form equation (1) in the above equation. Then,

$\begin{align}

  & \dfrac{3}{2}\left( {{\cos }^{2}}A+{{\sin }^{2}}A \right) \\

 & \Rightarrow \dfrac{3}{2} \\

\end{align}$

Now, from the above result, we can say that ${{\cos }^{2}}A+{{\cos }^{2}}\left( A+\dfrac{\pi }{3} \right)+{{\cos }^{2}}\left( A-\dfrac{\pi }{3} \right)=\dfrac{3}{2}$ .

Thus, $L.H.S=R.H.S$.

Hence Proved.

Note: Here, the student should first understand what we have to prove in the question and then proceed in a stepwise manner while solving. For making the simplification process smooth, we should also try to make use of trigonometric ratios like $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ and $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$ for making equations that will help us further in the solution. Moreover, the formulas like $\cos \left( A+B \right)$ and $\cos \left( A-B \right)$ should be applied correctly with proper signs, values and avoid making calculation mistakes while solving the problem.