Why does this equality hold?

$\sin x - \sin y = 2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$.

My professor was saying that since

(i) $\sin(A+B)=\sin A \cos B+ \sin B \cos A$


(ii) $\sin(A-B) = \sin A \cos B - \sin B \cos A$

we just let $A=\frac{x+y}{2}$ and $B=\frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated

  • $\begingroup$ Let A and B be as you defined. Then $sin(A+B)=sin(\frac{x+y}{2}+\frac{x-y}{2})$. Evaluate this and use the given identities. $\endgroup$ – Newman Apr 2 at 3:02
  • $\begingroup$ After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii) $\endgroup$ – R_D Apr 2 at 3:03

The main trick is here:

\begin{align} \color{red} {x = {x+y\over2} + {x-y\over2}}\\[1em] \color{blue}{y = {x+y\over2} - {x-y\over2}} \end{align}

(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)

Substituting the right-hand sides for $\color{red}x$ and $\color{blue}y,\,$ you will obtain

\begin{align} \sin \color{red} x - \sin \color{blue }y = \sin \left(\color{red}{{x+y\over2} + {x-y\over2} }\right) - \sin \left(\color{blue }{{x+y\over2} - {x-y\over2}} \right) \\[1em] \end{align}

All the rest is then only a routine calculation:

\begin{align} \require{enclose} &= \sin \left({x+y\over2}\right) \cos\left( {x-y\over2} \right) + \sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)\\[1em] &-\left[\sin \left({x+y\over2}\right) \cos\left( {x-y\over2} \right) - \sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)\right]\\[3em] &= \enclose{updiagonalstrike}{\sin \left({x+y\over2}\right) \cos\left( {x-y\over2} \right)} + \sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)\\[1em] &-\enclose{updiagonalstrike}{\sin \left({x+y\over2}\right) \cos\left( {x-y\over2} \right)} + \sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right) \\[3em] &=2\sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)\\ \end{align}


Following your professor's advice, let $A=\frac{x+y}{2}$, $B=\frac{x-y}{2}$. Then $$x=A+B\\y=A-B$$So the LHS of your equation becomes $$\sin(A+B)-\sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2\cos A\sin B$ as required.


Following your notation, let $A=\dfrac{x+y}{2}$ and $B=\dfrac{x-y}{2}$. Note that $A+B=x$ and $A-B=y$.

Now, $\sin x=\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin y=\sin(A-B)=\sin A\cos B - \cos A\sin B$ from your professor's advice.

To get the LHS, $\sin x-\sin y = 2\cos A\sin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.


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