Question

If $\sin 2a+\sin 2b=\dfrac{1}{2}$ and $\cos 2a+\cos 2b=\dfrac{3}{2}$ then ${{\cos }^{2}}\left( a-b \right)$ is equal to
A. $\dfrac{3}{8}$
B. $\dfrac{5}{8}$
C. 0
D. $\dfrac{5}{4}$

Answer 1

Hint: To solve this question we will use the addition and subtraction formula of sine and cosine functions. We will first square the equations individually and then add both the obtained equations. Simplifying the obtained equation we will get the desired answer.

Complete step by step answer:
We have been given that $\sin 2a+\sin 2b=\dfrac{1}{2}$ and $\cos 2a+\cos 2b=\dfrac{3}{2}$.
We have to find the value of ${{\cos }^{2}}\left( a-b \right)$.
Now, let us consider first equation then we will get
$\Rightarrow \sin 2a+\sin 2b=\dfrac{1}{2}$
Now, squaring both sides of the equation we will get
\[\Rightarrow {{\left( \sin 2a+\sin 2b \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}\]
Now, simplifying the above obtained equation we will get
\[\Rightarrow {{\sin }^{2}}2a+{{\sin }^{2}}2b+2\sin 2a.\sin 2b=\dfrac{1}{4}..........(i)\]
Now, let us consider second equation then we will get
$\Rightarrow \cos 2a+\cos 2b=\dfrac{3}{2}$
Now, squaring both sides of the equation we will get
\[\Rightarrow {{\left( \cos 2a+\cos 2b \right)}^{2}}={{\left( \dfrac{3}{2} \right)}^{2}}\]
Now, simplifying the above obtained equation we will get
\[\Rightarrow {{\cos }^{2}}2a+{{\cos }^{2}}2b+2\cos 2a.\cos 2b=\dfrac{9}{4}..........(ii)\]
Now, adding equation (i) and (ii) we will get
\[\Rightarrow {{\sin }^{2}}2a+{{\sin }^{2}}2b+2\sin 2a.\sin 2b+{{\cos }^{2}}2a+{{\cos }^{2}}2b+2\cos 2a.\cos 2b=\dfrac{9}{4}+\dfrac{1}{4}\]
Now, grouping the terms we will get
\[\Rightarrow \left( {{\sin }^{2}}2a+{{\cos }^{2}}2a \right)+\left( {{\sin }^{2}}2b+{{\cos }^{2}}2b \right)+2\left( \sin 2a.\sin 2b+\cos 2a.\cos 2b \right)=\dfrac{9}{4}+\dfrac{1}{4}\]
Now, we know that \[{{\sin }^{2}}a+{{\cos }^{2}}a=1\] and \[\sin a.\sin b+\cos a.\cos b=\cos \left( a-b \right)\]
Now, substituting the values and simplifying the obtained equation we will get
\[\begin{align}
  & \Rightarrow 1+1+2\cos 2\left( a-b \right)=\dfrac{9+1}{4} \\
 & \Rightarrow 2+2\cos 2\left( a-b \right)=\dfrac{10}{4} \\
 & \Rightarrow 2\cos 2\left( a-b \right)=\dfrac{5}{2}-2 \\
 & \Rightarrow 2\cos 2\left( a-b \right)=\dfrac{5-4}{2} \\
 & \Rightarrow 2\cos 2\left( a-b \right)=\dfrac{1}{2} \\
 & \Rightarrow \cos 2\left( a-b \right)=\dfrac{1}{4} \\
\end{align}\]
Now, we know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$
Now, applying the above formula to the obtained equation we will get
\[\Rightarrow 2{{\cos }^{2}}\left( a-b \right)-1=\dfrac{1}{4}\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow 2{{\cos }^{2}}\left( a-b \right)=\dfrac{1}{4}+1 \\
 & \Rightarrow 2{{\cos }^{2}}\left( a-b \right)=\dfrac{1+4}{4} \\
 & \Rightarrow {{\cos }^{2}}\left( a-b \right)=\dfrac{5}{8} \\
\end{align}\]

So, the correct answer is “Option B”.

Note: The possibility of mistake is while solving the equations. Students may solve the square formula as \[\Rightarrow {{\sin }^{2}}4a+{{\sin }^{2}}4b\] and so on. But 2a is the angle and sine is the function so the square is not applied to angle. So be careful while applying the formula else we get the incorrect answer.