Question

Prove that ${{\left( \cos A+\cos B \right)}^{2}}+{{\left( \sin A-\sin B \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{A+B}{2} \right)$.

Answer 1

Hint: For solving this problem, first we consider the left- hand side and expand it by using algebraic identities of ${{\left( a+b \right)}^{2}}\text{ and }{{\left( a-b \right)}^{2}}$. Now, we try to obtain a relationship in terms of angle (A +B) by applying suitable identities and expansion related to $\sin \theta \text{ and }\cos \theta $. By doing so, we obtain the term given on the right-hand side of the expression.

Complete step-by-step solution -
According to the problem statement, we are given expression ${{\left( \cos A+\cos B \right)}^{2}}+{{\left( \sin A-\sin B \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{A+B}{2} \right)$. First, we consider the left-hand of the expression and try to expand it by using the algebraic identities which can be stated as:
$\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align}$
So, \[{{\left( \cos A+\cos B \right)}^{2}}\] can be expanded as \[{{\cos }^{2}}A+{{\cos }^{2}}B+2\left( \cos A \right)\left( \cos B \right)\] and \[{{\left( \sin A-\sin B \right)}^{2}}\] can be expanded as \[{{\sin }^{2}}A+{{\sin }^{2}}B-2\left( \sin A \right)\left( \sin B \right)\]. Putting these values in left-hand side, we get
\[\begin{align}
  & \Rightarrow {{\left( \cos A+\cos B \right)}^{2}}+{{\left( \sin A-\sin B \right)}^{2}} \\
 & \Rightarrow {{\cos }^{2}}A+{{\cos }^{2}}B+2\left( \cos A \right)\left( \cos B \right)+{{\sin }^{2}}A+{{\sin }^{2}}B-2\left( \sin A \right)\left( \sin B \right) \\
\end{align}\]
Now, we rearrange the terms to separate A and B as,
\[\Rightarrow \left( {{\cos }^{2}}A+{{\sin }^{2}}A \right)+\left( {{\cos }^{2}}B+{{\sin }^{2}}B \right)+2\left( \cos A \right)\left( \cos B \right)-2\left( \sin A \right)\left( \sin B \right)\]
Now, by using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get
\[\begin{align}
  & \Rightarrow 1+1+2\left( \cos A \right)\left( \cos B \right)-2\left( \sin A \right)\left( \sin B \right) \\
 & \Rightarrow 2+2\left( \cos A \right)\left( \cos B \right)-2\left( \sin A \right)\left( \sin B \right) \\
 & \Rightarrow 2\left[ 1+\left( \cos A \right)\left( \cos B \right)-\left( \sin A \right)\left( \sin B \right) \right] \\
\end{align}\]
Now, by using the identity \[\left( \cos A \right)\left( \cos B \right)-\left( \sin A \right)\left( \sin B \right)=\cos \left( A+B \right)\], we get
\[\Rightarrow 2\left[ 1+\left( \cos A \right)\left( \cos B \right)-\left( \sin A \right)\left( \sin B \right) \right]=2\left( 1+\cos \left( A+B \right) \right)\]
Again, by using the identity $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$, we get
$\Rightarrow 2\left( 1+\cos \left( A+B \right) \right)=4{{\cos }^{2}}\left( \dfrac{A+B}{2} \right)$
So, the above obtained result is the same as given in the right-hand side of the problem expression. Hence, we proved the equivalence of left and right-hand sides.

Note: This problem can be alternatively solved by expanding $\sin A-\sin B\text{ and }\cos A+\cos B$ using the angle to multiplication property which can be stated as $\left[ \sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)\text{ and }\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) \right]$. Now, by squaring the above part individually and then adding, we obtain the same result as obtained above.