Chapter 5.PS, Problem 1PS

To determine

To write:

Each of all the other trigonometric functions of $\theta$ in terms of $\cos \theta$.

Answer to Problem 1PS

Solution:

All the other trigonometric functions of $\theta$ in terms of $\cos \theta$ are:

$\sin \theta =\pm \sqrt{1-{\cos }^2\theta },\tan \theta =\frac{\pm \sqrt{1-{\cos }^2\theta }}{\cos \theta },\csc \theta =\frac{1}{\pm \sqrt{1-{\cos }^2\theta }},\cot \theta =\frac{\cos \theta }{\pm \sqrt{1-{\cos }^2\theta }},$ and

$sec\theta =\frac{1}{\cos \theta }$.

Explanation of Solution

Formula:

Reciprocal identities:

$\begin{array}{l}\sin u=\frac{1}{\csc u},\cos u=\frac{1}{\sec u},\tan u=\frac{1}{\cot u}\\ \csc u=\frac{1}{\sin u},\sec u=\frac{1}{\cos u},\cot u=\frac{1}{\tan u}\end{array}$,

Quotient Identities:

$\tan u=\frac{\sin u}{\cos u},\cot u=\frac{\cos u}{\sin u}$

Pythagorean Identities:

$\begin{array}{rll}{\sin }^{2}u+{\cos }^{2}u& =& 1\\ 1+{\tan }^{2}u& =& {\sec }^{2}u\\ 1+{\cot }^{2}u& =& {\csc }^{2}u\end{array}$

Calculation:

Consider the provided trigonometric function $\cos \theta$. All the other trigonometric functions has to be expressed in terms of $\cos \theta$.

Use the Pythagorean identity ${\sin }^{2}u+{\cos }^{2}u=1$ to calculate $\sin \theta$. So,

${\sin }^2\theta +{\cos }^2\theta =1$

Subtract ${\cos }^2\theta$ from each side of the equation.

$\begin{array}{rll}{\sin }^2\theta +{\cos }^2\theta -{\cos }^2\theta & =& 1-{\cos }^2\theta \\ {\sin }^2\theta & =& 1-{\cos }^2\theta \end{array}$

Take square root on both the sides of the equation.

$\begin{array}{rll}\sqrt{{\sin }^2\theta }& =& \sqrt{1-{\cos }^2\theta }\\ \sin \theta & =& \pm \sqrt{1-{\cos }^2\theta }\end{array}$

Now, use the reciprocal identity $\csc u=\frac{1}{\sin u}$ and substitute $\sin \theta =\pm \sqrt{1-{\cos }^2\theta }$ to calculate $\csc \theta$.

$\begin{array}{rll}\csc \theta & =& \frac{1}{\sin \theta }\\ & =& \frac{1}{\pm \sqrt{1-{\cos }^2\theta }}\end{array}$

By the reciprocal identity $\sec u=\frac{1}{\cos u}$, $\sec \theta$ is calculated in terms of $\cos \theta$ as,

$\sec \theta =\frac{1}{\cos \theta }$

Use the quotient identity $\tan u=\frac{\sin u}{\cos u}$ and substitute $\sin \theta =\pm \sqrt{1-{\cos }^2\theta }$ to calculate $\tan \theta$. So,

$\tan \theta =\pm \frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }$

By the reciprocal identity $\cot u=\frac{1}{\tan u}$, $\cot \theta$ is calculated in terms of $\cos \theta$ as,

$\begin{array}{rll}\cot \theta & =& \frac{1}{\tan \theta }\\ & =& \frac{1}{\pm \frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }}\\ & =& \pm \frac{\cos \theta }{\sqrt{1-{\cos }^2\theta }}\end{array}$

Thus, all the other trigonometric functions are expressed in terms of $\cos \theta$.