2
$\begingroup$

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

and

(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

$\endgroup$
  • $\begingroup$ 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
9
$\begingroup$

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}

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.