Question

If $\pi < \theta < \dfrac{{3\pi }}{2}$ then the expression $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$ is equal to
(A) $2$
(B) $2 + 4\sin \theta$
(C) $2 - 4\sin \theta$
(D) 0

Answer 1

Use the double angle formula for cosine function, i.e. $\cos 2x = 1 - 2{\sin ^2}x = 2{\cos ^2} - 1$ for expression the ${\sin ^4}\theta$ in terms of $\cos 2\theta$. After that again use it to change $4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$ into the form $2\left( {1 + \cos \left( {\dfrac{\pi }{2} - \theta } \right)} \right)$ . Now transform it further to get the expression in similar trigonometric ratio.

Complete step-by-step answer:
In this problem, we are given an angle $'\theta '$ which lies in the interval $\left( {\pi ,\dfrac{{3\pi }}{2}} \right)$ . And there is a trigonometric expression $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$ given in terms of angle $\theta$ . Using the trigonometric properties and identities, we need to find the value of this expression.
According to the cosine double angle formula, we have:
$\cos 2x = 1 - 2{\sin ^2}x = 2{\cos ^2} - 1$ …………….(i)
Let’s change the ${\sin ^4}x$ term inside the radical sign in the form of $\cos 2x$. Therefore, by further transforming the above equation we can rewrite it as:
$\Rightarrow \cos 2x = 1 - 2{\sin ^2}x \Rightarrow 2{\sin ^2}x = 1 - \cos 2x$
On squaring both sides, we will have:
$\Rightarrow {\left( {2{{\sin }^2}x} \right)^2} = {\left( {1 - \cos 2x} \right)^2} \Rightarrow 4{\sin ^4}x = 1 + {\cos ^2}2x - 2\cos 2x$ ............(ii)
Again using (i), and expressing $\cos 2x$ in terms of ${\cos ^2}x$ as:
$\Rightarrow \cos 2x = 2{\cos ^2}x - 1 \Rightarrow 2{\cos ^2}x = 1 + \cos 2x$ ...............(iii)
Now let’s use these relations (ii) and (iii) in the given expression:
$\Rightarrow \sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right) = \sqrt {\left( {1 + {{\cos }^2}2\theta - 2\cos 2\theta } \right) + {{\sin }^2}2\theta } + 2\left( {1 + \cos \left( {\dfrac{\pi }{2} - \theta } \right)} \right)$
We know the identity ${\sin ^2}x + {\cos ^2}x = 1$ and $\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x$ . After using these identities in the above expression, we get:
$\Rightarrow \sqrt {1 + 1 - 2\cos 2\theta } + 2\left( {1 + \sin \theta } \right) = \sqrt {2 - 2\cos 2\theta } + 2 + 2\sin \theta$
Now let’s again use the relation (i) in the radical sign, and we will get:
$\Rightarrow \sqrt {2 - 2\cos 2\theta } + 2 + 2\sin \theta = \sqrt {2 - 2\left( {1 - 2{{\sin }^2}\theta } \right)} + 2 + 2\sin \theta = \sqrt {4{{\sin }^2}\theta } + 2 + 2\sin \theta$
Now, we just need to find the square root of $4{\sin ^2}\theta$ to solve the above expression. As we know that the range of the square-root function or exponential function is non-negative real numbers.
Therefore, we get: ${\left( {4{{\sin }^2}\theta } \right)^{\dfrac{1}{2}}} = \left| {2\sin \theta } \right|$
Here, the sign of $\sin \theta$ will determine the sign of square root of $4{\sin ^2}\theta$ . For an angle lying in the interval $\left( {\pi ,\dfrac{{3\pi }}{2}} \right)$ , the sine function will always give a negative value. So, to satisfy the range of square-root function, the absolute function should be resolved with a negative sign.
$\Rightarrow {\left( {4{{\sin }^2}\theta } \right)^{\dfrac{1}{2}}} = \left| {2\sin \theta } \right| = - 2\sin \theta$
Thus, our expression becomes:
$\Rightarrow \sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right) = \sqrt {4{{\sin }^2}\theta } + 2 + 2\sin \theta = - 2\sin \theta + 2 + 2\sin \theta = 2$
Therefore, we can conclude: $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right) = 2$ for $\theta \in \left( {\pi ,\dfrac{{3\pi }}{2}} \right)$
Hence, the option (A) is the correct answer.

Note: For questions like this, the knowledge about the signs of different ratios in different quadrants of angle plays a crucial role. An alternate approach is to break the given expression into two parts: $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta }$ and $4{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\theta }{2}} \right)$ . Now solve the first part using the formula for double angle of sine and solve the second part using the double angle formula for cosine. Now after combining you will get the required answer.