Question

Prove the following expression:
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta$

Answer 1

Hint: First of all take the LHS of the given expression and write it as ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta ={{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}$. Now, use the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ to expand it. Further use ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and proceed to simplify further to prove the desired result.

Complete step-by-step solution -
In this question, we have to prove that
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta$
First of all, let us take the LHS of the expression given in the question, we get,
$E={{\sin }^{6}}\theta +{{\cos }^{6}}\theta$
We know that ${{\left( {{a}^{2}} \right)}^{3}}={{a}^{6}}$. By using this in the above expression, we get,
$E={{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}$
We know that ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$. So by considering $a={{\sin }^{2}}\theta$ and $b={{\cos }^{2}}\theta$ and using this formula in the above expression, we get,
$E=\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\left[ {{\left( {{\sin }^{2}}\theta \right)}^{2}}+{{\left( {{\cos }^{2}}\theta \right)}^{2}}-{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right]$
$E=\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\left[ {{\sin }^{4}}\theta +{{\cos }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right]$
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. By using this in the above equation, we get,
$E=1.\left[ {{\sin }^{4}}\theta +{{\cos }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right]$
We know that ${{\left( {{a}^{2}} \right)}^{2}}={{a}^{4}}$. By using this in the above equation, we get,
$E={{\left( {{\sin }^{2}}\theta \right)}^{2}}+{{\left( {{\cos }^{2}}\theta \right)}^{2}}-{{\sin }^{2}}\theta {{\cos }^{2}}\theta$
We know that
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$
So by considering $a={{\sin }^{2}}\theta$ and $b={{\cos }^{2}}\theta$ and using this formula in the above expression, we get,
$E={{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta$
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. By using this in the above equation, we get,
$E=1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta$
$E=\left[ 1-{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left( 2+1 \right) \right]$
$E=\left[ 1-{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left( 2+1 \right) \right]$
$E=1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta$
So, LHS = RHS
Hence proved
So, we have proved that
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta$

Note: In this type of question, students must try to write the terms of the higher power in the combination of lower powers like we have written ${{\sin }^{6}}\theta ={{\left( {{\sin }^{2}}\theta \right)}^{3}}$ and ${{\sin }^{4}}\theta ={{\left( {{\sin }^{2}}\theta \right)}^{2}}$. In this way, we can easily use identities like ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ or ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ etc. These identities should be mentioned as there is extensive use of these identities in trigonometry as well as algebra.