If \[\cos \alpha + \cos \beta = \dfrac{3}{2}\] and \[\sin \alpha + \sin \beta = \dfrac{1}{2}\] and \[\theta \] is the arithmetic mean of \[\alpha ,\beta \], then \[\sin 2\theta + \cos 2\theta = \]
A) \[\dfrac{3}{5}\]
B) \[\dfrac{7}{5}\]
C) \[\dfrac{4}{5}\]
D) \[\dfrac{8}{5}\]

Hint: First of all calculate the equation for \[\theta \] using the given condition as, \[\theta = \dfrac{{\alpha + \beta }}{2}\] also we need to remember some basic formula such as \[\sin \alpha + \sin \beta = 2\sin (\dfrac{{\alpha + \beta }}{2})\cos (\dfrac{{\alpha - \beta }}{2})\] and \[\cos \alpha + \cos \beta = 2\cos (\dfrac{{\alpha + \beta }}{2})\cos (\dfrac{{\alpha - \beta }}{2})\] apply this both in the above equations and then we can replace \[\theta = \dfrac{{\alpha + \beta }}{2}\] in the equation and then divide both the equations in order to eliminate \[\cos (\dfrac{{\alpha - \beta }}{2})\]. Thus, finally the trigonometric equation will be obtained and simplify it to obtain the value of \[\sin \theta ,\cos \theta \] then we find the terms of \[\sin 2\theta ,\cos 2\theta \] as it is \[2\sin \theta \cos \theta ,2{\cos ^2}\theta - 1\] respectively, hence finally put the value and our required answer will be obtained.

Complete step by step solution: As the given equations are \[\cos \alpha + \cos \beta = \dfrac{3}{2}\]and \[\sin \alpha + \sin \beta = \dfrac{1}{2}\]
So, we can apply the half angle formula in both of the given equations as
\[\cos \alpha + \cos \beta = 2\cos (\dfrac{{\alpha + \beta }}{2})\cos (\dfrac{{\alpha - \beta }}{2}) = \dfrac{3}{2}\]and \[\sin \alpha + \sin \beta = 2\sin (\dfrac{{\alpha + \beta }}{2})\cos (\dfrac{{\alpha - \beta }}{2}) = \dfrac{1}{2}\]
As it is given \[\theta \] is the arithmetic mean of \[\alpha ,\beta \], we get \[\theta = \dfrac{{\alpha + \beta }}{2}\],
Now, replace \[\theta = \dfrac{{\alpha + \beta }}{2}\], in the above equations.
\[2\sin (\theta )\cos (\dfrac{{\alpha - \beta }}{2}) = \dfrac{1}{2}\]and \[2\cos (\theta )\cos (\dfrac{{\alpha - \beta }}{2}) = \dfrac{3}{2}\]
Hence, on dividing both the equations we can obtained the value of \[\tan \theta \] as,
\[\tan \theta = \dfrac{1}{3}\]
Now, as \[\tan \theta = \dfrac{{\sin (\theta )}}{{\cos (\theta )}} = \dfrac{1}{3}\]
So, the values of the \[\sin \theta ,\cos \theta \] can be calculated as,
\[
  \sin (\theta ) = \dfrac{1}{{\sqrt {{1^2} + {3^2}} }} = \dfrac{1}{{\sqrt {10} }} \\
  \cos (\theta ) = \dfrac{3}{{\sqrt {{1^2} + {3^2}} }} = \dfrac{3}{{\sqrt {10} }} \\
 \]
Now, simplify the given equations \[\sin 2\theta + \cos 2\theta \]as \[\sin 2\theta = 2\sin \theta \cos \theta \] and \[\cos 2\theta = 2{\cos ^2}\theta - 1\], we get,
\[\sin (2\theta ) + \cos (2\theta ) = 2\sin \theta \cos \theta + 2{\cos ^2}\theta - 1\]
Now, put the values in the above equation as per obtained earlier,
So,
\[
   = 2\sin \theta \cos \theta + 2{\cos ^2}\theta - 1 \\
   = 2(\dfrac{1}{{\sqrt {10} }})(\dfrac{3}{{\sqrt {10} }}) + 2(\dfrac{3}{{\sqrt {10} }})(\dfrac{3}{{\sqrt {10} }}) - 1 \\
 \]
On simplifying the above equation, we get,
\[
   = \dfrac{6}{{10}} + \dfrac{{18}}{{10}} - 1 \\
   = \dfrac{{24}}{{10}} - 1 \\
 \]
On taking LCM we get,
\[
   = \dfrac{{24 - 10}}{{10}} \\
   = \dfrac{{14}}{{10}} \\
 \]
On simplification we get,
\[ = \dfrac{7}{5}\]
Hence, \[\sin 2\theta + \cos 2\theta = \]\[\dfrac{7}{5}\].

Hence, option (B) is the correct answer.

Note: These kind of question are special in mathematics. The reason being, it involves the concept of two different topics. One is trigonometry and the other one is Arithmetic mean. This is very common with trigonometry. It can be clubbed with any topic and make the question complex. For us, as students, we need to stick to our concepts and process.