Question

Prove the following expression:
$\cos (a+b)-\cos \left( a-b \right)=-2\sin a\times \sin b$

Answer 1

Hint: We know the formulas \[\left( \cos (a+b)=\cos a\times \cos b-\sin a\times \sin b \right)\] and \[\left( \cos (a-b)=\cos a\times \cos b+\sin a\times \sin b \right)\]. So, we can simplify L.H.S. using the formula of \[\cos (a+b)\] and \[\cos (a-b)\].

Complete step-by-step answer:
We have to prove that, $\cos (a+b)-\cos \left( a-b \right)=-2\sin a\times \sin b$
Let us consider L.H.S. first, that is, $\cos (a+b)-\cos \left( a-b \right)$
As we know the formula that, \[\left( \cos (a+b)=\cos a\times \cos b-\sin a\times \sin b \right)......\left( i \right)\] and \[\left( \cos (a-b)=\cos a\times \cos b+\sin a\times \sin b \right)......\left( ii \right)\]
So, we can put the values of $\cos (a+b)$ and $\cos (a-b)$ in L.H.S to simplify it.
Therefore, we can write L.H.S as
$\Rightarrow \cos (a+b)-\cos \left( a-b \right)$
$\Rightarrow \left( \cos a\times \cos b-\sin a\times \sin b \right)-\left( \cos a\times \cos b+\sin a\times \sin b \right)$
Opening brackets to further simplify, so we get L.H.S. as
$\Rightarrow \cos a\times \cos b-\sin a\times \sin b-\cos a\times \cos b-\sin a\times \sin b$
In this, we can notice that it contains two terms of $\cos a\times \cos b$ with opposite signs and two terms of $\sin a\times \sin b$ with the same signs. And we know that terms with the same sign get added and terms with opposite signs get subtracted.
Therefore, we will get L.H.S. as
\[\Rightarrow -2\sin a\times \sin b\]
= R.H.S.
Hence, we have proved that $\cos (a+b)-\cos \left( a-b \right)=-2\sin a\times \sin b$.

Note: Another alternate method that can be used is:
Let us consider, \[x=a+b\] and \[y=a-b\]
Therefore, we can write, \[\left( \dfrac{x+y}{2} \right)=\left( \dfrac{\left( a+b \right)+\left( a-b \right)}{2} \right)\]
Which implies, \[\left( \dfrac{x+y}{2} \right)=\left( \dfrac{a+b+a-b}{2} \right)\]
\[\Rightarrow \left( \dfrac{x+y}{2} \right)=\left( \dfrac{2a}{2} \right)\]
\[\Rightarrow \left( \dfrac{x+y}{2} \right)=\left( a \right)......\left( iii \right)\]
Similarly, we can write \[\left( \dfrac{x-y}{2} \right)=\left( b \right)......\left( iv \right)\]
Now, we are considering L.H.S. first.
So, L.H.S. = \[\cos \left( a+b \right)-\cos \left( a-b \right)\]
As we have considered \[a+b=x\] and \[a-b=y\]
Therefore, L.H.S. can be written as
\[\Rightarrow \cos \left( x \right)-\cos \left( y \right)\]
As we know, \[\left[ \cos \left( x \right)-\cos \left( y \right)=-2\sin \left( \dfrac{x+y}{2} \right)\times \sin \left( \dfrac{x-y}{2} \right) \right]\]
So, we may get L.H.S. as
\[\Rightarrow -2\sin \left( \dfrac{x+y}{2} \right)\times \sin \left( \dfrac{x-y}{2} \right)\]
Now, we will put the values of \[\left( \dfrac{x+y}{2} \right)\] and \[\left( \dfrac{x-y}{2} \right)\] from (iii) and (iv)
So, we will get L.H.S. as
\[\Rightarrow -2\sin \left( a \right)\times \sin \left( b \right)\]
Hence proved.