Question

Prove that the expression ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta $ is valid.

Answer 1

Hint: While answering this question we need to prove the expression ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta $ . For that we will use ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ from the formula ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

Complete step-by-step solution:
Now let us consider from the question, we need to prove that ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta $.
As we know that ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ from the formula ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ .
 By using the above expression and expanding the L.H.S of the given expression in the question we will have ${{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{3}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)$ .
From the basic concept we know that the formula stating the relation between $\sin \theta $ and $\cos \theta $ is given as ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
 By using the above relation in the expression we have we will get the expression $1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta $ .
If we compare the R.H.S and the expression we have now, we come to the conclusion that both are completely the same.
Hence, we can say that the L.H.S of the given equation is equal to the R.H.S of the respective equation.
Hence, it is proved that the expression ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta $ is valid.

Note: For any proof we need to use the L.H.S and derive the expression to R.H.S for those derivations we can use 3 trigonometric identities ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta $ using the Pythagoras identities that states that in a right-angle triangle, the square of the hypotenuse side that is the side opposite to the right angle is equal to the sum of the squares of the adjacent and opposite sides of the right-angle triangle.