Chapter 5.4, Problem 67E

To determine

To write the expression $\csc \left(\frac{3\pi }{2}+\theta \right)$ as a trigonometric function of only $\theta$ and to verify the answer graphically

Answer to Problem 67E

  $-\sec \theta$

Explanation of Solution

Given:

Expression: $\csc \left(\frac{3\pi }{2}+\theta \right)$

Formula Used:

  $\begin{array}{l}\csc A=\frac{1}{\sin A}\\ \sin \left(A+B\right)=\sin A\cos B+\cos A\sin B\end{array}$

Calculation:

  $\begin{array}{l}\Rightarrow \csc \left(\frac{3\pi }{2}+\theta \right)\\ \Rightarrow \frac{1}{\sin \left(\frac{3\pi }{2}+\theta \right)}[\because \csc A=\frac{1}{\sin A}]\\ \Rightarrow \frac{1}{\sin \frac{3\pi }{2}\cos \theta +\cos \frac{3\pi }{2}\sin \theta }[\because \sin \left(A+B\right)=\sin A\cos B+\cos A\sin B]\end{array}$

But,

  $\begin{array}{l}\cos \frac{3\pi }{2}=0\\ \sin \frac{3\pi }{2}=-1\end{array}$

Plugging the above values in the above expression,

  $\begin{array}{l}\Rightarrow \frac{1}{\left(-1\right)\cos \theta +\left(0\right)\sin \theta }\\ \Rightarrow \frac{1}{-\cos \theta +0}\\ \Rightarrow \frac{1}{-\cos \theta }\\ \Rightarrow -\sec \theta \end{array}$

Hence, $\csc \left(\frac{3\pi }{2}+\theta \right)=-\sec \theta .$

Verification:

Let $f(x)=\csc \left(\frac{3\pi }{2}+\theta \right)\text{and}g(x)=-\sec \theta$

Now, using the graphing utility to graph the above functions,

Graph of f (x):

  

Graph of g(x):

  

Interpretation:

From observing both graphs, it is clear that, they are equal to each other.

Hence, $\csc \left(\frac{3\pi }{2}+\theta \right)=-\sec \theta .$