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.