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,
$\begin{aligned} & \text{A) }13-4{{\cos }^{6}}\theta \\\ & \text{B) }13-4{{\cos }^{4}}\theta +2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ & \text{B) }13-4{{\cos }^{2}}\theta +6{{\cos }^{4}}\theta \\\ & \text{B) }13-4{{\cos }^{2}}\theta +6{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ \end{aligned}$

Answer 1

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 ${{\left( \sin \theta -\cos \theta \right)}^{2}}$.

& 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{aligned}$$Since, $${{\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{aligned} & 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{aligned}$$$$\begin{aligned} & 3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =3-12\sin \theta \cos \theta +12{{\sin }^{2}}\theta {{\cos }^{2}}\theta +6 +12\sin \theta \cos \theta \\\ & \text{ }+4{{\sin }^{6}}\theta \\\ \end{aligned}$$$$3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =3+12{{\sin }^{2}}\theta {{\cos }^{2}}\theta +6 +4{{\sin }^{6}}\theta $$ $$3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+12{{\sin }^{2}}\theta {{\cos }^{2}}\theta +4{{\sin }^{6}}\theta .....(1)$$ 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 $$\begin{aligned} & 3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+12{{\sin }^{2}}\theta {{\cos }^{2}}\theta +4{{\sin }^{2}}\theta -8{{\cos }^{2}}\theta {{\sin }^{2}}\theta \\\ & +4{{\cos }^{4}}\theta {{\sin }^{2}}\theta \\\ \end{aligned}$$ $$3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+4{{\sin }^{2}}\theta {{\cos }^{2}}\theta +4{{\sin }^{2}}\theta +4{{\cos }^{4}}\theta {{\sin }^{2}}\theta $$ $$3{{\left( \sin \theta -\cos \theta \right)}^{4}}+6{{\left( \sin \theta +\cos \theta \right)}^{2}}+4{{\sin }^{6}}\theta =9+4{{\sin }^{2}}\theta \left( {{\cos }^{2}}\theta +1+{{\cos }^{4}}\theta \right)$$ 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.