Question

For any \[\theta \in \left( \dfrac{\pi }{4},\dfrac{\pi }{2} \right),\text{ the expression }3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta \]
is,
$\begin{align}
  & \text{A) }13-4{{\cos }^{6}}\theta \\
 & \text{B) }13-4{{\cos }^{4}}\theta +2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\
 & \text{B) }13-4{{\cos }^{2}}\theta +6{{\cos }^{4}}\theta \\
 & \text{B) }13-4{{\cos }^{2}}\theta +6{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\
\end{align}$

Answer 1

Hint: In this problem, we will first express \[{{\left( \sin \theta -\cos \theta \right)}^{2}}\text{ and }{{\left( \sin \theta +\cos \theta \right)}^{2}}\] using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.we will also use \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Complete step by step answer:
The given expression
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =3{{\left( {{\left( \sin \theta -\cos \theta \right)}^{2}} \right)}^{2}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta \]
Now we will use ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ to express \[{{\left( \sin \theta +\cos \theta \right)}^{2}}\] and \[{{\left( \sin \theta -\cos \theta \right)}^{2}}\].
\[\begin{align}
  & 3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =3{{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta \right)}^{2}}+6({{\sin }^{2}}\theta +{{\cos }^{2}}\theta \\
 & +2\sin \theta \cos \theta )+4{{\sin }^{6}}\theta \\
\end{align}\]Since, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =3{{\left( 1-2\sin \theta \cos \theta \right)}^{2}}+6(1 +2\sin \theta \cos \theta )+4{{\sin }^{6}}\theta \]
Again by using ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ to express \[{{\left( 1-2\sin \theta \cos \theta \right)}^{2}}\],
we get
\[\begin{align}
  & 3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =3\left( 1-4\sin \theta \cos \theta +4{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)+6(1 +2\sin \theta \cos \theta ) \\
 & \text{ }+4{{\sin }^{6}}\theta \\
\end{align}\]\[\begin{align}
  & 3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =3-12\sin \theta \cos \theta +12{{\sin }^{2}}\theta {{\cos }^{2}}\theta +6 +12\sin \theta \cos \theta \\
 & \text{ }+4{{\sin }^{6}}\theta \\
\end{align}\]\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =3+12{{\sin }^{2}}\theta {{\cos }^{2}}\theta +6 +4{{\sin }^{6}}\theta \]
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+12{{\sin }^{2}}\theta {{\cos }^{2}}\theta +4{{\sin }^{6}}\theta .....(1)\]

Now we will first simplify \[4{{\sin }^{6}}\theta \] and then use in equation (1)
\[4{{\sin }^{6}}\theta =4{{\sin }^{2}}\theta \cdot {{\sin }^{4}}\theta \]
Since , \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]
\[4{{\sin }^{6}}\theta =4\left( 1-{{\cos }^{2}}\theta \right)\cdot {{\sin }^{4}}\theta \]
\[\Rightarrow 4{{\sin }^{6}}\theta =4{{\sin }^{4}}\theta -4{{\cos }^{2}}\theta {{\sin }^{4}}\theta \cdot \]
\[\Rightarrow 4{{\sin }^{6}}\theta =4{{\sin }^{2}}\theta {{\sin }^{2}}\theta -4{{\cos }^{2}}\theta {{\sin }^{2}}\theta {{\sin }^{2}}\theta \cdot \]
Again by using \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \] we get
\[\Rightarrow 4{{\sin }^{6}}\theta =4\left( 1-{{\cos }^{2}}\theta \right){{\sin }^{2}}\theta -4{{\cos }^{2}}\theta {{\sin }^{2}}\theta \left( 1-{{\cos }^{2}}\theta \right)\cdot \]
\[\Rightarrow 4{{\sin }^{6}}\theta =4{{\sin }^{2}}\theta -4{{\cos }^{2}}\theta {{\sin }^{2}}\theta -4{{\cos }^{2}}\theta {{\sin }^{2}}\theta +4{{\cos }^{4}}\theta {{\sin }^{2}}\theta \cdot \]
\[\Rightarrow 4{{\sin }^{6}}\theta =4{{\sin }^{2}}\theta -8{{\cos }^{2}}\theta {{\sin }^{2}}\theta +4{{\cos }^{4}}\theta {{\sin }^{2}}\theta .....(2)\]

Using equation (2) in equation (1), we get
\[\begin{align}
  & 3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+12{{\sin }^{2}}\theta {{\cos }^{2}}\theta +4{{\sin }^{2}}\theta -8{{\cos }^{2}}\theta {{\sin }^{2}}\theta \\
 & +4{{\cos }^{4}}\theta {{\sin }^{2}}\theta \\
\end{align}\]
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+4{{\sin }^{2}}\theta {{\cos }^{2}}\theta +4{{\sin }^{2}}\theta +4{{\cos }^{4}}\theta {{\sin }^{2}}\theta \]
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+4{{\sin }^{2}}\theta \left( {{\cos }^{2}}\theta +1+{{\cos }^{4}}\theta \right)\]
Again by using \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]
we get
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+4\left( 1-{{\cos }^{2}}\theta \right)\left( {{\cos }^{2}}\theta +1+{{\cos }^{4}}\theta \right)\]
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+4\left( {{\cos }^{2}}\theta +1+{{\cos }^{4}}\theta -{{\cos }^{4}}\theta -{{\cos }^{2}}\theta -{{\cos }^{6}}\theta \right)\]\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+4\left( 1-{{\cos }^{6}}\theta \right)\]
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+4-4{{\cos }^{6}}\theta \]
\[3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =13-4{{\cos }^{6}}\theta \]

So, the correct answer is “Option A”.

Note: In this problem, we can check that we option is correct by substituting the values of theta (option by elimination). Since the given expression is true for any \[\theta \in \left( \dfrac{\pi }{4},\dfrac{\pi }{2} \right)\] that is true for any one of the options.