Questions & Answers

Question

Answers

Answer

Verified

128.7k+ views

\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]

\[{(a + b)^2} = {a^2} + {b^2} + 2ab\]

\[{a^2} + {b^2} = {(a + b)^2} - 2ab\]

\[{a^4} - {b^4} = ({a^2} - {b^2})({a^2} + {b^2})\]

\[{a^2} - {b^2} = (a - b)(a + b)\]

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 \[{\sin ^8}\theta - {\cos ^8}\theta = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}\theta {\cos ^2}\theta )\]