Question

1 –54 Use the guidelines of this section to sketch the curve.

39. \(y = \frac{{\sin x}}{{1 + \cos x}}\)

Short answer

Graph of the given curve are,

Step-by-step solution

Steps for sketching a curve

There are following terms needed to examine forSketching a Graphfor any given Function:

1. Find the Domainof the Function.
2. Calculate the Intercepts.
3. Check forSymmetricity.
4. Find theAsymptotes.
5. Intervals ofIncrease and Decrease.
6. Evaluate theMaxima and Minimaof the function.
7. ExamineConcavityand the Point of Inflection.
8. Sketch the Graph.

(A) Determine Domain

The given function is \(y = \frac{{\sin x}}{{1 + \cos x}}\).

Simplify the given function by rationalizing the denominator.

\(\begin{array}{c}y = \frac{{\sin x}}{{1 + \cos x}} \cdot \frac{{1 - \cos x}}{{1 - \cos x}}\\ = \frac{{\sin x\left( {1 - \cos x} \right)}}{{\left( {1 - {{\cos }^2}x} \right)}}\\ = \frac{{\sin x\left( {1 - \cos x} \right)}}{{{{\sin }^2}x}}\\ = \frac{{1 - \cos x}}{{\sin x}}\\ = \frac{1}{{\sin x}} - \frac{{\cos x}}{{\sin x}}\\ = \csc x - \cot x\end{array}\)

The domain of the given function is the set of all real numbers except the odd multiples of \(\pi \), that is \(\left( {2n + 1} \right)\pi \), where \(n\) is an integer.

(B) Determine Intercepts

Find \(y\)-intercepts by substituting 0 for \(x\) into \(y = \frac{{\sin x}}{{1 + \cos x}}\).

\(\begin{array}{c}y = \frac{{\sin 0}}{{1 + \cos 0}}\\ = \frac{0}{2}\\ = 0\end{array}\)

So, \(y\)-intercept is \(\left( {0,0} \right)\).

Find \(x\)-intercepts by substituting 0 for \(y\) into \(y = \frac{{\sin x}}{{1 + \cos x}}\).

\(\begin{array}{c}0 = \frac{{\sin x}}{{1 + \cos x}}\\\sin x = 0\\x = 2n\pi \end{array}\)

So, \(x\)-intercept is \(\left( {2n\pi ,0} \right)\), where \(n\) is an integer.

(C) Determine symmetry

For the given function \(y = \frac{{\sin x}}{{1 + \cos x}}\), \(f\left( { - x} \right) = - x\), so the function is symmetric about origin. The function has a period of \(2\pi \), because \(\frac{{\sin \left( {2\pi + x} \right)}}{{1 + \cos \left( {2\pi + x} \right)}} = \frac{{\sin x}}{{1 + \cos x}}\).

(D) Find Asymptotes 

There is no horizontal asymptote because infinity is not included in the domain.

Determine \(\mathop {\lim }\limits_{x \to n{\pi ^ + }} y\) and \(\mathop {\lim }\limits_{x \to n{\pi ^ - }} y\) for Vertical asymptotes, where \(n\) is a odd integer.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to n{\pi ^ + }} y = \mathop {\lim }\limits_{x \to n{\pi ^ + }} \left( {\frac{{\sin x}}{{1 + \cos x}}} \right)\\ = - \infty \end{array}\)

\(\begin{array}{c}\mathop {\lim }\limits_{x \to n{\pi ^ + }} y = \mathop {\lim }\limits_{x \to n{\pi ^ + }} \left( {\frac{{\sin x}}{{1 + \cos x}}} \right)\\ = \infty \end{array}\)

So, the Vertical asymptote is \(x = n\pi \) for odd integer \(n\).

(E) Find Intervals of Increase or Decrease 

Find the first derivative of the given function with respect to \(x\).

\(\begin{array}{c}y' = \frac{d}{{dx}}\left( {\frac{{\sin x}}{{1 + \cos x}}} \right)\\ = \frac{{\left( {1 + \cos x} \right) \cdot \cos x - \sin x\left( { - \sin x} \right)}}{{{{\left( {1 + \cos x} \right)}^2}}}\\ = \frac{{1 + \cos x}}{{{{\left( {1 + \cos x} \right)}^2}}}\\ = \frac{1}{{1 + \cos x}}\end{array}\)

Here, \(y' = 0\) does not exist. And \(y' > 0\) for all \(x\) except the odd multiples of \(\pi \)

Draw a table for the interval of increasing and decreasing.

Interval	\(y'\)	Behaviour of \(y\)
\(\left( {\left( {2k - 1} \right)\pi ,\left( {2k + 1} \right)\pi } \right)\)	+	Increasing

(F) Find Local Minimum and Maximum values 

From the obtained table and the condition of local maxima and minima, it can be said that, there is no local maximum/minimum value because sign of \(y'\) does changes.

(G) Determine Concavity and points of Inflection 

Find \(y''\).

\(\begin{array}{c}y'' = \frac{d}{{dx}}\left( {\frac{1}{{1 + \cos x}}} \right)\\ = \frac{1}{{{{\left( {1 + \cos x} \right)}^2}}} \cdot \sin x\\ = \frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}\end{array}\)

For \(y'' = 0\), \(x = 2n\pi \) and it is not defined at \(x = \left( {2k - 1} \right)\pi ,\left( {2k + 1} \right)\pi \).

Draw a table for concavity for different intervals.

Interval	Sign of \(y''\)	Behaviour of \(y\)
\(\left( {\left( {2k - 1} \right)\pi ,2k\pi } \right)\)	-	Concave downward
\(x = 2k\pi \)	0	Inflection
\(\left( {2k\pi ,\left( {2k + 1} \right)\pi } \right)\)	+	Concave upward

The inflection point is at \(x = 2k\pi \).

And, \(y\left( {2k\pi } \right) = 0\). So, the inflection point is \(\left( {2k\pi ,0} \right)\). Here, \(k\) is an integer.

(H) Draw Graph

By using all the obtained information from step 2 to 8, draw the graph of the given function.