 QUESTION

# Prove that $\left( {{\sin }^{8}}\theta -{{\cos }^{8}}\theta \right)=\left( {{\sin }^{2}}\theta -{{\cos }^{2}}\theta \right)\left( 1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)$ .

Hint: For solving this question we will simplify the term written on the left-hand side and then try to prove it equal to the term written on the right-hand side of the equation with the help of some trigonometric identities and the whole square formula.

Given:
We have to prove that $\left( {{\sin }^{8}}\theta -{{\cos }^{8}}\theta \right)=\left( {{\sin }^{2}}\theta -{{\cos }^{2}}\theta \right)\left( 1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)$

Now, before we proceed to the simplification of the term on the left-hand side. We should see the formula written below. We will be using it in our simplification. The formula is as follows:

We know that, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ .

Then, ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab...........\left( 1 \right)$ .

We know that, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right).......\left( 2 \right)$ .

We will use one of the very basic and important formulas of trigonometry here. The formula is written below:

${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1..............\left( 3 \right)$

Now, we will prove the result by simplifying the term on the L.H.S.

We have, $\left( {{\sin }^{8}}\theta -{{\cos }^{8}}\theta \right)$ . Now using the result of the equation (2). Then,

\begin{align} & \left( {{\sin }^{8}}\theta -{{\cos }^{8}}\theta \right)=\left( {{\left( {{\sin }^{4}}\theta \right)}^{2}}-{{\left( {{\cos }^{4}}\theta \right)}^{2}} \right) \\ & \Rightarrow \left( {{\sin }^{4}}\theta -{{\cos }^{4}}\theta \right)\left( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta \right) \\ & \Rightarrow \left( {{\left( {{\sin }^{2}}\theta \right)}^{2}}-{{\left( {{\cos }^{2}}\theta \right)}^{2}} \right)\left( {{\left( {{\sin }^{2}}\theta \right)}^{2}}+{{\left( {{\cos }^{2}}\theta \right)}^{2}} \right) \\ & \Rightarrow \left( {{\sin }^{2}}\theta -{{\cos }^{2}}\theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\left( {{\left( {{\sin }^{2}}\theta \right)}^{2}}+{{\left( {{\cos }^{2}}\theta \right)}^{2}} \right) \\ \end{align}

Now, using the result from equation (3) and equation (2) to simplify the above expression. Then

\begin{align} & \left( {{\sin }^{2}}\theta -{{\cos }^{2}}\theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\left( {{\left( {{\sin }^{2}}\theta \right)}^{2}}+{{\left( {{\cos }^{2}}\theta \right)}^{2}} \right) \\ & \Rightarrow \left( {{\sin }^{2}}\theta -{{\cos }^{2}}\theta \right)\left( {{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right) \\ & \Rightarrow \left( {{\sin }^{2}}\theta -{{\cos }^{2}}\theta \right)\left( {{\left( 1 \right)}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right) \\ & \Rightarrow \left( {{\sin }^{2}}\theta -{{\cos }^{2}}\theta \right)\left( 1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right) \\ \end{align}

Now, from the above calculation $\left( {{\sin }^{8}}\theta -{{\cos }^{8}}\theta \right)=\left( {{\sin }^{2}}\theta -{{\cos }^{2}}\theta \right)\left( 1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)$ .

Thus, we have proved that term on the left-hand side is equal to the term on the right-hand side.

Hence, proved.

Note: Although the problem was easy to solve, the student should apply the formula correctly in a proper sequence and solve the question without any calculation mistake to prove the result easily.