Question

Prove the following expression: ${{\left( \cos x-\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}=4{{\sin }^{2}}\left( \dfrac{x-y}{2} \right)$.

Answer 1

Hint: To solve this question, we should know a few trigonometric identities like, $\cos 2\theta =1-2{{\sin }^{2}}\theta $ and $\cos a\cos b+\sin a\sin b=\cos \left( a-b \right)$ . We should also know a few algebraic identities too like, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. By using these respectively, we can prove the expression given in the question.

Complete step-by-step answer:
In the given question, we have been asked to prove the expression, which is, ${{\left( \cos x-\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}=4{{\sin }^{2}}\left( \dfrac{x-y}{2} \right)$. For doing that, we will first consider the left hand side or the LHS, that is, ${{\left( \cos x-\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}$. We know that${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, so by applying this in the LHS, we will get, $\begin{align}
  & {{\cos }^{2}}x+{{\cos }^{2}}y-2\left( \cos x \right)\left( \cos y \right)+{{\sin }^{2}}x+{{\sin }^{2}}y-2\left( \sin x \right)\left( \sin y \right) \\
 & \Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x+{{\cos }^{2}}y+{{\sin }^{2}}y-2\left[ \left( \cos x \right)\left( \cos y \right)+\left( \sin x \right)\left( \sin y \right) \right] \\
\end{align}$
Now, we also know that, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Now, we can apply that in the above equation, so we will get the LHS as,
$\begin{align}
  & 1+1-2\left[ \left( \cos x \right)\left( \cos y \right)+\left( \sin x \right)\left( \sin y \right) \right] \\
 & \Rightarrow 2-2\left[ \left( \cos x \right)\left( \cos y \right)+\left( \sin x \right)\left( \sin y \right) \right] \\
\end{align}$
We also know that $\left( \cos x \right)\left( \cos y \right)+\left( \sin x \right)\left( \sin y \right)=\cos \left( x-y \right)$. By applying that in the above equation, we get the LHS as, $2-2\left[ \cos \left( x-y \right) \right]\Rightarrow 2\left[ 1-\cos \left( x-y \right) \right]$. We also know that $\cos \alpha =1-2{{\sin }^{2}}\left( \dfrac{\alpha }{2} \right)$, which implies that, $2{{\sin }^{2}}\left( \dfrac{\alpha }{2} \right)=1-\cos \alpha $ or we can also say that, $4{{\sin }^{2}}\left( \dfrac{\alpha }{2} \right)=2\left( 1-\cos \alpha \right)$. So, by substituting the same in the LHS, we get, $2\left[ 1-\cos \left( x-y \right) \right]=4{{\sin }^{2}}\left( \dfrac{x-y}{2} \right)$, which is equal to the RHS of the expression given in the question. Hence, we have proved the expression given in the question, ${{\left( \cos x-\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}=4{{\sin }^{2}}\left( \dfrac{x-y}{2} \right)$.

Note: While solving this type of questions, we need to remember that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Sometimes, we do get stuck at a few points while proving the expressions, in that case we can also consider the RHS and simplify that side till we reach the values given in the LHS to prove the given expression.