Question

Prove the following: $\dfrac{{{\cos }^{3}}x-{{\sin }^{3}}x}{\cos x-\sin x}=\dfrac{1}{2}\left( 2+\sin 2x \right)$.

Answer 1

Hint: To solve this question, we need to know a few trigonometric relations like, ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ and $\sin 2x=2\sin x\cos x$. We should also know a few algebraic relations like, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. By using these relations, we can prove the required relation in the question easily.

Complete step-by-step answer:

In this question, we have been asked to prove, $\dfrac{{{\cos }^{3}}x-{{\sin }^{3}}x}{\cos x-\sin x}=\dfrac{1}{2}\left( 2+\sin 2x \right)$. To prove this relation, we will first consider the left hand side or the LHS of the relation, which is, $=\dfrac{{{\cos }^{3}}x-{{\sin }^{3}}x}{\cos x-\sin x}$ . We know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. So, we can apply that in the LHS, to get the LHS as the following,

$\dfrac{\left( \cos x-\sin x \right)\left( {{\cos }^{2}}x+{{\sin }^{2}}x+\cos x\sin x \right)}{\left( \cos x-\sin x \right)}$

We can see here that $\left( \cos x-\sin x \right)$ is common in both the numerator and the denominator, so we can cancel them. After cancelling the like terms, we get the LHS as,
$={{\cos }^{2}}x+{{\sin }^{2}}x+\cos x\sin x$

We know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$, so, by applying that in the LHS side, we get the LHS as,$=1+\cos x\sin x$

We also know that $2\cos x\sin x=\sin 2x$ or we can say $\cos x\sin x=\dfrac{\sin 2x}{2}$. So, by applying it in the LHS side, we get the LHS as, $=1+\dfrac{\sin 2x}{2}$
.
Now, we will take the LCM for further simplification. By taking the LCM, we get the LHS as,

$\begin{align}

  & =\dfrac{2+\sin 2x}{2} \\

 & =\dfrac{1}{2}\left( 2+\sin 2x \right) \\

\end{align}$

Which is the same as the right hand side or the RHS of the relation given in the question, that is, $\dfrac{1}{2}\left( 2+\sin 2x \right)$. Hence, we have proved the relation given in the question, which is, $\dfrac{{{\cos }^{3}}x-{{\sin }^{3}}x}{\cos x-\sin x}=\dfrac{1}{2}\left( 2+\sin 2x \right)$.

Note: The possible mistakes the students can make in this question is that they may write ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$, while in a hurry and they may end up with the wrong result as the correct expression to be used is ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.