Question

The maximum value of the expression $ \dfrac{1}{{{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta } $ is

A. 0

B. 1

C. 2

D. 3

E. 4

Answer 1

Hint: We need to find the minimum value of $ {{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta $ to find the maximum value of the expression $ \dfrac{1}{{{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta } $ . First, simplify the expression using trigonometric formulas. Then we try to convert the equation into an equation of singular ratio using the formula of $ \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ . Then we apply the condition of minimum value to get the answer.

Complete step by step answer:

First, we try to find the simplified form of the equation $ {{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta $ .

We have the identity theorem of $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ .

We convert the equation as

$ \Rightarrow {{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta $

$=\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+\dfrac{3}{2}\left( 2\sin \theta \cos \theta \right)+4{{\cos }^{2}}\theta $

$ =1+\dfrac{3}{2}\left( 2\sin \theta \cos \theta \right)+2\left( 2{{\cos }^{2}}\theta \right) $

Now we apply the theorems of $ \sin 2\theta =2\sin \theta \cos \theta $ and $ 1+\cos 2\theta =2{{\cos }^{2}}\theta $.

$ \Rightarrow 1+\dfrac{3}{2}\left( 2\sin \theta \cos \theta \right)+2\left( 2{{\cos }^{2}}\theta \right) $

$ =1+\dfrac{3}{2}\left( \sin 2\theta \right)+2\left( 1+\cos 2\theta \right) $

$ =3+\dfrac{3\sin 2\theta }{2}+2\cos 2\theta $

To find the maximum value of the expression $ \dfrac{1}{{{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta } $ , we need to find the minimum value of $ {{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta $ which is equal to $ 3+\dfrac{3\sin 2\theta }{2}+2\cos 2\theta $ .

Now we try to convert the equation into one particular ratio.

We will take $\dfrac{5}{2} $ common from sin and cos terms and get

$ 3+\dfrac{3\sin 2\theta }{2}+2\cos 2\theta =3+\dfrac{5}{2}\left( \dfrac{3}{5}\sin 2\theta +\dfrac{4}{5}\cos 2\theta \right) $ .

We assume $ \cos \alpha =\dfrac{3}{5} $ which gives $ \sin \alpha =\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5} $ .

The equation becomes $ 3+\dfrac{5}{2}\left( \dfrac{3}{5}\sin 2\theta +\dfrac{4}{5}\cos 2\theta \right)=3+\dfrac{5}{2}\left( \sin 2\theta \cos \alpha +\cos 2\theta \sin \alpha \right) $ .

Now we apply $ \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ . We assume $ A=2\theta ;B=\alpha $ .

\[3+\dfrac{5}{2}\left( \sin 2\theta \cos \alpha +\cos 2\theta \sin \alpha \right)=3+\dfrac{5}{2}\sin \left( 2\theta +\alpha \right)\].

Now we know that for any value of $ x\in \mathbb{R} $, the minimum value of $ \sin x $ is -1.

This means the minimum value of \[\sin \left( 2\theta +\alpha \right)\] is \[-1\]. We will find the minimum value of \[3+\dfrac{5}{2}\sin \left( 2\theta +\alpha \right)\] by putting minimum value of \[\sin \left( 2\theta +\alpha \right)\] as \[-1\].

So,\[3+\dfrac{5}{2}\sin \left( 2\theta +\alpha \right)=\]\[3+\dfrac{5}{2}\times(-1)\]

\[=\dfrac{1}{2}\]

Therefore, the minimum value of $ 3+\dfrac{3\sin 2\theta }{2}+2\cos 2\theta $ which is equal to $ {{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta $ , is $ \dfrac{1}{2} $ .

The maximum value of the expression $ \dfrac{1}{{{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta } $ is $ \dfrac{1}{\dfrac{1}{2}}=2 $ .

The correct option is (C).

Note:

The general formula of finding minimum and maximum value of equation types like $ a\sin \alpha +b\cos \alpha $ is $ -\sqrt{{{a}^{2}}+{{b}^{2}}}\le \left( a\sin \alpha +b\cos \alpha \right)\le \sqrt{{{a}^{2}}+{{b}^{2}}} $ . For any values of $ x\in \mathbb{R} $ , minimum and maximum values of ratios like $ \sin x,\cos x $ is $ -1\le \sin x,\cos x\le 1 $ .