Question

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

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Hint: Here in this question some trigonometric identities and algebraic identities will get used. They are mentioned below: -
${\sin ^2}\theta + {\cos ^2}\theta = 1$
${(a + b)^2} = {a^2} + {b^2} + 2ab$
${a^2} + {b^2} = {(a + b)^2} - 2ab$
${a^4} - {b^4} = ({a^2} - {b^2})({a^2} + {b^2})$
${a^2} - {b^2} = (a - b)(a + b)$

We will convert the L.H.S (left hand side) part equal to R.H.S (right hand side) using some trigonometric identities and algebraic identities.
First of all we will split the power 8 like this ${a^8} - {b^8} = {({a^4})^2} - {({b^4})^2}$ so that we can apply further identity.
$\Rightarrow {({\sin ^4}\theta )^2} - {({\cos ^4}\theta )^2}$
Now applying identity ${a^2} - {b^2} = (a - b)(a + b)$ we will get
$\Rightarrow ({\sin ^4}\theta - {\cos ^4}\theta )({\sin ^4}\theta + {\cos ^4}\theta )$
Now we will split power 4 so that we can apply identity.
$\Rightarrow ({\sin ^4}\theta - {\cos ^4}\theta )[{({\sin ^2}\theta )^2} + {({\cos ^2}\theta )^2}]$
Now we will apply identity${a^2} + {b^2} = {(a + b)^2} - 2ab$we will get
$\Rightarrow ({\sin ^4}\theta - {\cos ^4}\theta )[{({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta ]$
Now we will apply identity${a^4} - {b^4} = ({a^2} - {b^2})({a^2} + {b^2})$we will get
$\Rightarrow ({\sin ^2}\theta - {\cos ^2}\theta )({\sin ^2}\theta + {\cos ^2}\theta )[{({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta ]$
Now we will apply trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$and we will get
$\Rightarrow ({\sin ^2}\theta - {\cos ^2}\theta )[{({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta ]$
Now we will apply identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$again and we will get
$\Rightarrow ({\sin ^2}\theta - {\cos ^2}\theta )[{(1)^2} - 2{\sin ^2}\theta {\cos ^2}\theta ]$
$\therefore ({\sin ^2}\theta - {\cos ^2}\theta )[1 - 2{\sin ^2}\theta {\cos ^2}\theta ]$
Hence it is proved that ${\sin ^8}\theta - {\cos ^8}\theta = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}\theta {\cos ^2}\theta )$

Note: Students are likely to make mistakes in relating algebraic identities with trigonometric functions but they can be applied easily by only carefully observing the degrees. With the use of algebraic identities questions can be solved with very less complexity. Further this question can be solved alternatively by proofing the R.H.S (right hand side) part equal to L.H.S (left hand side). Same steps will take place but in reverse. In the previous case we eliminate the term as if to simplify the equation. Now in an alternate method we will just add more terms so that we can make power 8. That method will be more complex than the method we have used so the mentioned method must be followed to avoid any discrepancies.