Question

Prove the following trigonometric equation:
${{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$

Answer 1

Hint:As you can see the L.H.S of the given equation then you will find that $\cos x+\cos y$ is in the form of $\cos C+\cos D$ identity and $\sin x-\sin y$ is in the form of $\sin C-\sin D$ then apply these identities in the respective sine and cosine expression and then simplify.

Complete step-by-step answer:
The equation given in the question that we have to prove is:
${{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$
We are going to simplify L.H.S of the above equation and then try to resolve into R.H.S.
${{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}$
In the above expression, we can apply $\cos C+\cos D$ identity in $\cos x+\cos y$ as follows:
$\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
$\cos x+\cos y=2\cos \dfrac{x+y}{2}\cos \dfrac{x-y}{2}$
In the above expression, we can apply $\sin C-\sin D$ identity in $\sin x-\sin y$ as follows:
$\begin{align}
  & \sin C-\sin D=2\cos \dfrac{C+D}{2}\sin \dfrac{C-D}{2} \\
 & \sin x-\sin y=2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2} \\
\end{align}$
Now, substituting these value of trigonometric expressions in sine and cosine in${{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}$ we get,
$\begin{align}
  & {{\left( 2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) \right)}^{2}}+{{\left( 2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right) \right)}^{2}} \\
 & =4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right){{\cos }^{2}}\left( \dfrac{x-y}{2} \right)+4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right){{\sin }^{2}}\left( \dfrac{x-y}{2} \right) \\
\end{align}$
Taking $4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$ as common from the above expression we get,
$4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)\left( {{\cos }^{2}}\left( \dfrac{x-y}{2} \right)+{{\sin }^{2}}\left( \dfrac{x-y}{2} \right) \right)$
We know the trigonometric identity that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ . You can see this identity in the above expression where $\theta =\dfrac{x-y}{2}$. So applying this identity in the above expression we get,
$\begin{align}
  & 4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)\left( 1 \right) \\
 & =4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right) \\
\end{align}$
The R.H.S of the given equation is $4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$.
From the simplification of L.H.S of the given equation, the answer we got is $4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$which is equal to R.H.S.
Hence, we have proved that L.H.S = R.H.S of the given equation.

Note: There is an alternative way of proving the above equation is as follows:
${{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$
We are going to simplify the L.H.S of the above equation.
${{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}$
Now, we are going to open the square of cosine and sine expressions in the above equation we get,
${{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y+{{\sin }^{2}}x+{{\sin }^{2}}y-2\sin x\sin y$
Rearranging the above expression we get,
${{\cos }^{2}}x+{{\sin }^{2}}x+{{\cos }^{2}}y+{{\sin }^{2}}y+2\cos x\cos y-2\sin x\sin y$
Now, we can use the identity of ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ in the above expression.
$1+1+2\left( \cos x\cos y-\sin x\sin y \right)$
We know that $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$. Using this cosine relation in the above equation we get,
$\begin{align}
  & 2+2\cos \left( x+y \right) \\
 & =2\left( 1+\cos \left( x+y \right) \right) \\
\end{align}$
From the double angle of cosine identity we know that $1+\cos 2\theta =2{{\cos }^{2}}\theta $. So in the above expression we can use the relation $x+y=2\theta $ . Using this double angle cosine identity in the above expression we get,
$\begin{align}
  & 2\left( 2{{\cos }^{2}}\left( \dfrac{x+y}{2} \right) \right) \\
 & =4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right) \\
\end{align}$
R.H.S of the given equation is $4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$.
And from the above simplification of L.H.S of the given equation, we are getting answer as$4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$ which is equal to R.H.S.
Hence, we have proved that L.H.S = R.H.S of the given equation.