Question

Simplify the following trigonometric expression.
 $\dfrac{\sin 2\theta }{1+\cos 2\theta }$.

Answer 1

Hint: To solve this problem, we should be aware about the basic properties of trigonometric terms involving multiple angles. Thus, we should know that-
$\begin{align}
  & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\
 & \sin 2\theta =2\sin \theta \cos \theta \\
\end{align}$
These expressions would be helpful in solving the above mentioned problem.

Complete step-by-step answer:
Before we begin solving the question, it is important to know about multiple angles in relation to trigonometry. Generally, to find the expression of $\sin 2\theta $, we express $2\theta =\theta +\theta $ and then use the property that $\sin (A+B)=\sin A\cos B+\cos A\sin B$. Thus, in this case, A=B=$\theta $. Thus, we have,
$\sin (\theta +\theta )=\sin \theta \cos \theta +\sin \theta \cos \theta $
$\sin (2\theta )=2\sin \theta \cos \theta $
We can similarly derive the formula for $\cos 2\theta $ by a similar procedure. We use the property that cos(A+B) = cosAcosB – sinAsinB. Thus, in this case, A=B=$\theta $. Thus, we have,
$\begin{align}
  & \cos (\theta +\theta )=\cos \theta \cos \theta -\sin \theta \sin \theta \\
 & \cos (2\theta )={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\
 & \cos (2\theta )={{\cos }^{2}}\theta -(1-{{\cos }^{2}}\theta ) \\
 & \cos (2\theta )=2{{\cos }^{2}}\theta -1 \\
\end{align}$

Thus, we make use of these properties to solve the above problem. Now, we have,
$\dfrac{\sin 2\theta }{1+\cos 2\theta }$
 \[\begin{align}
  & =\dfrac{2\sin \theta \cos \theta }{1+2{{\cos }^{2}}\theta -1} \\
 & =\dfrac{2\sin \theta \cos \theta }{2{{\cos }^{2}}\theta } \\
 & =\dfrac{\sin \theta }{\cos \theta } \\
 & =\tan \theta \\
\end{align}\]

Hence, the correct answer after simplification is tan$\theta $.

Note: An alternative way to solve this problem is to express denominator and numerator in terms of tan$\theta $ and then solve. Thus, we have,
=$\dfrac{\sin 2\theta }{1+\cos 2\theta }$
Now, to get expression in terms of tan$\theta $, we divide the numerator and denominator by cos$2\theta $, thus we get,
=$\dfrac{\tan 2\theta }{1+\sec 2\theta }$
Since, $\dfrac{\sin 2\theta }{\cos 2\theta }=\tan 2\theta $ and $\dfrac{1}{\cos 2\theta }=\sec 2\theta $.
By using the property that $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ and $\sec 2\theta =\dfrac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }$, we get,
 $\begin{align}
  & =\dfrac{\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }}{1+\dfrac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }} \\
 & \\
 & =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta +1+{{\tan }^{2}}\theta } \\
 & \\
 & =\dfrac{2\tan \theta }{2} \\
 & \\
 & =\tan \theta \\
\end{align}$

Thus, we were able to arrive at the same solution using a different technique. Thus, in question involving solving a trigonometric expression, there are generally multiple ways to arrive at the solution, however, at times it is generally beneficial to use one method over another due to its ease.