Question

Given $\left\{ \begin{align}
  & \sin x+\sin y=a \\
 & \cos x+\cos y=b \\
\end{align} \right.$, Calculate $\sin \left( \dfrac{x-y}{2} \right)$?

Answer 1

Hint: We first take the square value of the given expressions. We simplify them and add them to the identity formula of ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$. We find the simplified form in the sum and use the formula of \[\cos x\cos y+\sin x\sin y=\cos \left( x-y \right)\]. Then we use the submultiple formula of \[\cos \alpha =1-2{{\sin }^{2}}\dfrac{\alpha }{2}\]. We get $\sin \left( \dfrac{x-y}{2} \right)$ in expression. We simplify to find the solution.

Complete step by step solution:
We have been given two equations $\left\{ \begin{align}
  & \sin x+\sin y=a \\
 & \cos x+\cos y=b \\
\end{align} \right.$.
We take the squares of the equations and simplify them. We use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
$\begin{align}
  & {{\left( \sin x+\sin y \right)}^{2}}={{a}^{2}} \\
 & \Rightarrow {{\sin }^{2}}x+{{\sin }^{2}}y+2\sin x\sin y={{a}^{2}} \\
\end{align}$
$\begin{align}
  & {{\left( \cos x+\cos y \right)}^{2}}={{b}^{2}} \\
 & \Rightarrow {{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y={{b}^{2}} \\
\end{align}$
We now add them and get
$\begin{align}
  & {{\sin }^{2}}x+{{\sin }^{2}}y+2\sin x\sin y+{{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow \left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)+\left( {{\sin }^{2}}y+{{\cos }^{2}}y \right)+2\left( \sin x\sin y+\cos x\cos y \right)={{a}^{2}}+{{b}^{2}} \\
\end{align}$
We now use the identity theorem of ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$.
\[\begin{align}
  & \left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)+\left( {{\sin }^{2}}y+{{\cos }^{2}}y \right)+2\left( \sin x\sin y+\cos x\cos y \right)={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow 1+1+2\left( \sin x\sin y+\cos x\cos y \right)={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow \cos x\cos y+\sin x\sin y=\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2} \\
\end{align}\]
We have the associative formula of \[\cos x\cos y+\sin x\sin y=\cos \left( x-y \right)\].
Therefore, \[\cos x\cos y+\sin x\sin y=\cos \left( x-y \right)=\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2}\].
We now use the identity \[\cos \alpha =1-2{{\sin }^{2}}\dfrac{\alpha }{2}\].
Therefore, \[\cos \left( x-y \right)=1-2{{\sin }^{2}}\left( \dfrac{x-y}{2} \right)=\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2}\].
We now simplify to find the value of $\sin \left( \dfrac{x-y}{2} \right)$.
\[\begin{align}
  & 1-2{{\sin }^{2}}\left( \dfrac{x-y}{2} \right)=\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2} \\
 & \Rightarrow 2{{\sin }^{2}}\left( \dfrac{x-y}{2} \right)=1-\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2}=\dfrac{4-{{a}^{2}}-{{b}^{2}}}{2} \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{x-y}{2} \right)=\dfrac{4-{{a}^{2}}-{{b}^{2}}}{4} \\
 & \Rightarrow \sin \left( \dfrac{x-y}{2} \right)=\pm \sqrt{\dfrac{4-{{a}^{2}}-{{b}^{2}}}{4}}=\pm \dfrac{\sqrt{4-{{a}^{2}}-{{b}^{2}}}}{2} \\
\end{align}\]
The value of $\sin \left( \dfrac{x-y}{2} \right)$ is \[\pm \dfrac{\sqrt{4-{{a}^{2}}-{{b}^{2}}}}{2}\].

Note:
We can’t break the given expressions using the formulas $\sin x+\sin y=2\sin \dfrac{x+y}{2}\cos \dfrac{x-y}{2}$ and \[\cos x+\cos y=2\cos \dfrac{x+y}{2}\cos \dfrac{x-y}{2}\]. In that case we can only omit the \[\cos \dfrac{x-y}{2}\] and keep the angle of \[\dfrac{x+y}{2}\]. It is not helpful as we need the \[\cos \dfrac{x-y}{2}\] part for $\sin \left( \dfrac{x-y}{2} \right)$.