Question

Find $\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1$.
A. $2\sin 2\theta $
B. $2\cos 2\theta $
C. $\tan 2\theta $
D. $\cot 2\theta $

Answer 1

Hint: We first use multiple angle formulas and complete the summation with 1. We use the identity formulas like ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, \[2\sin \theta \cos \theta =\sin 2\theta \]. Then we take $4\left( \sin \theta +\cos \theta \right)$ common to cancel it out from the denominator.

Complete answer:
We have the trigonometric multiple angle formula where $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta $ and $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $.
We place these values in $\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1$ and simplify
$\begin{align}
  & \dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1 \\
 & =\dfrac{3\sin \theta -4{{\sin }^{3}}\theta -4{{\cos }^{3}}\theta +3\cos \theta }{\sin \theta +\cos \theta }+1 \\
 & =\dfrac{4\sin \theta -4{{\sin }^{3}}\theta -4{{\cos }^{3}}\theta +4\cos \theta }{\sin \theta +\cos \theta } \\
\end{align}$
Now we try to take 4 common and factorise the numerator.
$\begin{align}
  & 4\sin \theta -4{{\sin }^{3}}\theta -4{{\cos }^{3}}\theta +4\cos \theta \\
 & =4\left( \sin \theta +\cos \theta \right)-4\left( {{\sin }^{3}}\theta +{{\cos }^{3}}\theta \right) \\
\end{align}$
We use the cubic expansion as ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$.
So, we get ${{\sin }^{3}}\theta +{{\cos }^{3}}\theta =\left( \sin \theta +\cos \theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\sin \theta \cos \theta \right)$.
We know the identity value of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
So, ${{\sin }^{3}}\theta +{{\cos }^{3}}\theta =\left( \sin \theta +\cos \theta \right)\left( 1-\sin \theta \cos \theta \right)$.
We now again take common of $\left( \sin \theta +\cos \theta \right)$.
$\begin{align}
  & 4\left( \sin \theta +\cos \theta \right)-4\left( {{\sin }^{3}}\theta +{{\cos }^{3}}\theta \right) \\
 & =4\left( \sin \theta +\cos \theta \right)-4\left( \sin \theta +\cos \theta \right)\left( 1-\sin \theta \cos \theta \right) \\
 & =4\left( \sin \theta +\cos \theta \right)\left( 1-1+\sin \theta \cos \theta \right) \\
 & =4\sin \theta \cos \theta \left( \sin \theta +\cos \theta \right) \\
\end{align}$
The fraction form becomes \[\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1=\dfrac{4\sin \theta \cos \theta \left( \sin \theta +\cos \theta \right)}{\left( \sin \theta +\cos \theta \right)}=4\sin \theta \cos \theta \].
We know the identity formula of \[2\sin \theta \cos \theta =\sin 2\theta \].
So, \[4\sin \theta \cos \theta =2\left( 2\sin \theta \cos \theta \right)=2\sin 2\theta \].
Final solution is \[\dfrac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }+1=2\sin 2\theta \].
And hence the correct answer is option A.

Note:
It is important to remember that the condition to eliminate the $\left( \sin \theta +\cos \theta \right)$ from both denominator and numerator is $\left( \sin \theta +\cos \theta \right)\ne 0$. No domain is given for the variable $x$. The value of $\tan x\ne -1$ is essential. The simplified condition will be $x\ne n\pi -\dfrac{\pi }{4},n\in \mathbb{Z}$.