Question

Prove the following statement
${{\sin }^{8}}A-{{\cos }^{8}}A=\left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)\left( 1-2{{\sin }^{2}}A{{\cos }^{2}}A \right)$

Answer 1

Hint: To prove the statement, we need to have an idea about a few algebraic identities like, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ and also know that ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$. We should also remember a few trigonometric identities like, ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. By using these identities, we can prove the required relation.
Complete step-by-step answer:
In this question, we have been asked to prove that ${{\sin }^{8}}A-{{\cos }^{8}}A=\left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)\left( 1-2{{\sin }^{2}}A{{\cos }^{2}}A \right)$. For that, we will first consider the LHS of the given relation. So, we get,
\[LHS={{\sin }^{8}}A-{{\cos }^{8}}A\]
Now, we know that ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$, so we can write ${{\sin }^{8}}A={{\left( {{\sin }^{4}}A \right)}^{2}}$ and ${{\cos }^{8}}A={{\left( {{\cos }^{4}}A \right)}^{2}}$. Therefore, we get,
$LHS={{\left( {{\sin }^{4}}A \right)}^{2}}-{{\left( {{\cos }^{4}}A \right)}^{2}}$
Now, we know that, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. So, we can write the LHS as,
$LHS=\left[ \left( {{\sin }^{4}}A \right)-\left( {{\cos }^{4}}A \right) \right]\left[ \left( {{\sin }^{4}}A \right)+\left( {{\cos }^{4}}A \right) \right]$
Now, we know that, ${{\sin }^{4}}A$ and ${{\cos }^{4}}A$ can be written as ${{\left( {{\sin }^{2}}A \right)}^{2}}$ and ${{\left( {{\cos }^{2}}A \right)}^{2}}$, because ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$. Therefore, we can write the LHS as,
$LHS=\left[ {{\left( {{\sin }^{2}}A \right)}^{2}}-{{\left( {{\cos }^{2}}A \right)}^{2}} \right]\left[ {{\left( {{\sin }^{2}}A \right)}^{2}}+{{\left( {{\cos }^{2}}A \right)}^{2}} \right]$
Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, we can write ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$. Hence, we can write ${{\left( {{\sin }^{2}}A \right)}^{2}}+{{\left( {{\cos }^{2}}A \right)}^{2}}={{\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)}^{2}}-2{{\sin }^{2}}A{{\cos }^{2}}A$. Therefore we get the LHS as,
$LHS=\left[ {{\left( {{\sin }^{2}}A \right)}^{2}}-{{\left( {{\cos }^{2}}A \right)}^{2}} \right]\left[ {{\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)}^{2}}-2{{\sin }^{2}}A{{\cos }^{2}}A \right]$
Now, we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. So, we can write ${{\left( {{\sin }^{2}}A \right)}^{2}}-{{\left( {{\cos }^{2}}A \right)}^{2}}=\left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)$. Therefore, we get,
\[LHS=\left[ \left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right) \right]\left[ {{\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)}^{2}}-2{{\sin }^{2}}A{{\cos }^{2}}A \right]\]
Now, we know that, ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. So, we get the LHS as,
\[\begin{align}
  & LHS=\left[ \left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)\left( 1 \right) \right]\left[ {{\left( 1 \right)}^{2}}-2{{\sin }^{2}}A{{\cos }^{2}}A \right] \\
 & LHS=\left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)\left( 1-2{{\sin }^{2}}A{{\cos }^{2}}A \right) \\
\end{align}\]
And according to the given statement, we can say that the RHS is equal to \[\left( {{\sin }^{2}}A-{{\cos }^{2}}A \right)\left( 1-2{{\sin }^{2}}A{{\cos }^{2}}A \right)\].
Thus, we get $LHS=RHS$.
Hence proved.

Note: While solving this question, the students can think of starting from the RHS by multiplying the terms in both the brackets with each other, but after that step, they might get stuck as the solution becomes more complicated. Whereas, when we started from the LHS, while solving we had substituted ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ and that term gets removed from the multiplication and thus the solution became less complicated and easier to solve.