Question

How do you verify $2\cos \left( a+b \right)\sin \left( a-b \right)=\sin \left( 2a \right)-\sin \left( 2b \right)$?

Answer 1

Hint: In the above question, we have been given a trigonometric identity to prove. For proving it, we need to consider the left hand side of the given identity and apply the trigonometric identities $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ and $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$ to expand it. Then simplifying the expression obtained, and using the trigonometric identities $2\sin x\cos x=\sin 2x$ and ${{\cos }^{2}}x+{{\sin }^{2}}x=1$, we will obtain the expression present on the right hand side, and hence the given identity will be proved.

Complete step by step answer:
The identity given in the question is
\[2\cos \left( a+b \right)\sin \left( a-b \right)=\sin \left( 2a \right)-\sin \left( 2b \right).......(i)\]
Considering the term on the LHS, we have
\[\Rightarrow LHS=2\cos \left( a+b \right)\sin \left( a-b \right)........(ii)\]
Now, we know that
$\begin{align}
  & \cos \left( x+y \right)=\cos x\cos y-\sin x\sin y........(iii) \\
 & \sin \left( x-y \right)=\sin x\cos y-\cos x\sin y........(iv) \\
\end{align}$
Comparing the above identities with the equation (ii), we have $x=a$ and $y=b$. Substituting these in (iii) and (iv) we get
\[\begin{align}
  & \Rightarrow \cos \left( a+b \right)=\cos a\cos b-\sin a\sin b........(v) \\
 & \Rightarrow \sin \left( a-b \right)=\sin a\cos b-\cos a\sin b........(vi) \\
\end{align}\]
Substituting the equations (v) and (vi) in (i), we get
\[\Rightarrow LHS=2\left( \cos a\cos b-\sin a\sin b \right)\left( \sin a\cos b-\cos a\sin b \right)\]
Multiplying the two binomials, we get
\[\begin{align}
  & \Rightarrow LHS=2\left[ \cos a\cos b\left( \sin a\cos b-\cos a\sin b \right)-\sin a\sin b\left( \sin a\cos b-\cos a\sin b \right) \right] \\
 & \Rightarrow LHS=2\left[ \sin a\cos a{{\cos }^{2}}b-{{\cos }^{2}}a\sin b\cos b-{{\sin }^{2}}a\sin b\cos b+\sin a\cos a{{\sin }^{2}}b \right] \\
\end{align}\]
Taking $2$ inside, we have
\[\begin{align}
  & \Rightarrow LHS=2\sin a\cos a{{\cos }^{2}}b-2{{\cos }^{2}}a\sin b\cos b-2{{\sin }^{2}}a\sin b\cos b+2\sin a\cos a{{\sin }^{2}}b \\
 & \Rightarrow LHS=\left( 2\sin a\cos a \right){{\cos }^{2}}b-{{\cos }^{2}}a\left( 2\sin b\cos b \right)-{{\sin }^{2}}a\left( 2\sin b\cos b \right)+\left( 2\sin a\cos a \right){{\sin }^{2}}b......(vii) \\
\end{align}\]
Now, we know the trigonometric identity
$2\sin x\cos x=\sin 2x$
Putting $x=a$ and $x=b$ in the above identities, we get the equations
\[\begin{align}
  & 2\sin a\cos a=\sin \left( 2a \right)........(viii) \\
 & 2\sin b\cos b=\sin \left( 2b \right).........(ix) \\
\end{align}\]
Substituting the equations (viii) and (ix) in the equation (vii), we get
\[\begin{align}
& \Rightarrow LHS=\sin \left( 2a \right){{\cos }^{2}}b-{{\cos }^{2}}a\sin \left( 2b \right)-{{\sin }^{2}}a\sin \left( 2b \right)+\sin \left( 2a \right){{\sin }^{2}}b \\
& \Rightarrow LHS=\sin \left( 2a \right){{\cos }^{2}}b+\sin \left( 2a \right){{\sin }^{2}}b-{{\cos }^{2}}a\sin \left( 2b \right)-{{\sin }^{2}}a\sin \left( 2b \right) \\
\end{align}\]
Taking $\sin \left( 2a \right)$ common from the first two terms and $-\sin \left( 2b \right)$ common from the last two terms, we get
\[\Rightarrow LHS=\sin \left( 2a \right)\left( {{\cos }^{2}}b+{{\sin }^{2}}b \right)-\sin \left( 2b \right)\left( {{\cos }^{2}}a+{{\sin }^{2}}a \right)........(x)\]
Now, we know the identity
${{\cos }^{2}}x+{{\sin }^{2}}x=1$
Putting $x=a$ and $x=b$ in the above, we get
$\begin{align}
  & {{\cos }^{2}}a+{{\sin }^{2}}a=1........(xi) \\
 & {{\cos }^{2}}b+{{\sin }^{2}}b=1.........(xii) \\
\end{align}$
Substituting the equations (xi) and (xii) in the equation (x), we get
\[\Rightarrow LHS=\sin \left( 2a \right)-\sin \left( 2b \right)........(x)\]
From (i), we have
$RHS=\sin \left( 2a \right)-\sin \left( 2b \right)........(xi)$
Finally from the equations (x) and (xi) we conclude that
$LHS=RHS$
Hence, the given identity has been proved.

Note: To solve this question, we considered the left hand side of the given identity and with the help of the trigonometric identities which are known to us, we expanded the left hand side and manipulated it in the form of the right hand side. But instead of this if we considered the right hand side of the given identity; we could be able to prove it in just a single step. We just need to apply the trigonometric identity $\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)$ on the right hand side of the given identity, and we will get the expression manipulated in the form of the left hand side.