Question

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

Answer 1

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{aligned} & \sin x+\sin y=a \\\ & \cos x+\cos y=b \\\ \end{aligned} \right..
We take the squares of the equations and simplify them. We use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
$\begin{aligned} & {{\left( \sin x+\sin y \right)}^{2}}={{a}^{2}} \\\ & \Rightarrow {{\sin }^{2}}x+{{\sin }^{2}}y+2\sin x\sin y={{a}^{2}} \\\ \end{aligned}$
$\begin{aligned} & {{\left( \cos x+\cos y \right)}^{2}}={{b}^{2}} \\\ & \Rightarrow {{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y={{b}^{2}} \\\ \end{aligned}$
We now add them and get
$\begin{aligned} & {{\sin }^{2}}x+{{\sin }^{2}}y+2\sin x\sin y+{{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y={{a}^{2}}+{{b}^{2}} \\\ & \Rightarrow \left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)+\left( {{\sin }^{2}}y+{{\cos }^{2}}y \right)+2\left( \sin x\sin y+\cos x\cos y \right)={{a}^{2}}+{{b}^{2}} \\\ \end{aligned}$
We now use the identity theorem of ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$.

& \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{aligned}$$ We have the associative formula of $$\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{aligned} & 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{aligned}$$ The value of $\sin \left( \dfrac{x-y}{2} \right)$ is $$\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 $\sin \left( \dfrac{x-y}{2} \right)$.