# 7.4 The other trigonometric functions  (Page 4/14)

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## Using identities to evaluate trigonometric functions

1. Given $\text{\hspace{0.17em}}\mathrm{sin}\left(45°\right)=\frac{\sqrt{2}}{2},\mathrm{cos}\left(45°\right)=\frac{\sqrt{2}}{2},$ evaluate $\text{\hspace{0.17em}}\mathrm{tan}\left(45°\right).$
2. Given $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{5\pi }{6}\right)=\frac{1}{2},\mathrm{cos}\left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2},$ evaluate $\text{\hspace{0.17em}}\mathrm{sec}\left(\frac{5\pi }{6}\right).$

Because we know the sine and cosine values for these angles, we can use identities to evaluate the other functions.

1. $\begin{array}{ccc}\hfill \mathrm{tan}\left(45°\right)& =& \frac{\mathrm{sin}\left(45°\right)}{\mathrm{cos}\left(45°\right)}\hfill \\ & =& \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\hfill \\ & =& 1\hfill \end{array}$

2. $\begin{array}{ccc}\hfill \mathrm{sec}\left(\frac{5\pi }{6}\right)& =& \frac{1}{\mathrm{cos}\left(\frac{5\pi }{6}\right)}\hfill \\ & =& \frac{1}{-\frac{\sqrt{3}}{2}}\hfill \\ & =& \frac{-2\sqrt{3}}{1}\hfill \\ & =& \frac{-2}{\sqrt{3}}\hfill \\ & =& -\frac{2\sqrt{3}}{3}\hfill \end{array}$

Evaluate $\text{\hspace{0.17em}}\text{csc}\left(\frac{7\pi }{6}\right).$

$-2$

## Using identities to simplify trigonometric expressions

Simplify $\text{\hspace{0.17em}}\frac{\mathrm{sec}\text{\hspace{0.17em}}t}{\mathrm{tan}\text{\hspace{0.17em}}t}.$

We can simplify this by rewriting both functions in terms of sine and cosine.

By showing that $\text{\hspace{0.17em}}\frac{\mathrm{sec}\text{\hspace{0.17em}}t}{\mathrm{tan}\text{\hspace{0.17em}}t}\text{\hspace{0.17em}}$ can be simplified to $\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}t,$ we have, in fact, established a new identity.

$\frac{\mathrm{sec}\text{\hspace{0.17em}}t}{\mathrm{tan}\text{\hspace{0.17em}}t}=\mathrm{csc}\text{\hspace{0.17em}}t$

Simplify $\text{\hspace{0.17em}}\left(\mathrm{tan}\text{\hspace{0.17em}}t\right)\left(\mathrm{cos}\text{\hspace{0.17em}}t\right).$

$\mathrm{sin}t$

## Alternate forms of the pythagorean identity

We can use these fundamental identities to derive alternate forms of the Pythagorean Identity, $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1.\text{\hspace{0.17em}}$ One form is obtained by dividing both sides by $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t.$

$\begin{array}{ccc}\hfill \frac{{\mathrm{cos}}^{2}t}{{\mathrm{cos}}^{2}t}+\frac{{\mathrm{sin}}^{2}t}{{\mathrm{cos}}^{2}t}& =& \frac{1}{{\mathrm{cos}}^{2}t}\hfill \\ \hfill 1+{\mathrm{tan}}^{2}t& =& {\mathrm{sec}}^{2}t\hfill \end{array}$

The other form is obtained by dividing both sides by $\text{\hspace{0.17em}}{\mathrm{sin}}^{2}t.$

$\begin{array}{ccc}\hfill \frac{{\mathrm{cos}}^{2}t}{{\mathrm{sin}}^{2}t}+\frac{{\mathrm{sin}}^{2}t}{{\mathrm{sin}}^{2}t}& =& \frac{1}{{\mathrm{sin}}^{2}t}\hfill \\ \hfill {\mathrm{cot}}^{2}t+1& =& {\mathrm{csc}}^{2}t\hfill \end{array}$

## Alternate forms of the pythagorean identity

$1+{\mathrm{tan}}^{2}t={\mathrm{sec}}^{2}t$
${\mathrm{cot}}^{2}t+1={\mathrm{csc}}^{2}t$

## Using identities to relate trigonometric functions

If $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)=\frac{12}{13}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is in quadrant IV, as shown in [link] , find the values of the other five trigonometric functions.

We can find the sine using the Pythagorean Identity, $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1,$ and the remaining functions by relating them to sine and cosine.

The sign of the sine depends on the y -values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the y -values are negative, its sine is negative, $\text{\hspace{0.17em}}-\frac{5}{13}.$

The remaining functions can be calculated using identities relating them to sine and cosine.

If $\text{\hspace{0.17em}}\mathrm{sec}\left(t\right)=-\frac{17}{8}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}0 find the values of the other five functions.

As we discussed at the beginning of the chapter, a function that repeats its values in regular intervals is known as a periodic function. The trigonometric functions are periodic. For the four trigonometric functions, sine, cosine, cosecant and secant, a revolution of one circle, or $\text{\hspace{0.17em}}2\pi ,$ will result in the same outputs for these functions. And for tangent and cotangent, only a half a revolution will result in the same outputs.

Other functions can also be periodic. For example, the lengths of months repeat every four years. If $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ represents the length time, measured in years, and $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ represents the number of days in February, then $\text{\hspace{0.17em}}f\left(x+4\right)=f\left(x\right).$ This pattern repeats over and over through time. In other words, every four years, February is guaranteed to have the same number of days as it did 4 years earlier. The positive number 4 is the smallest positive number that satisfies this condition and is called the period. A period is the shortest interval over which a function completes one full cycle—in this example, the period is 4 and represents the time it takes for us to be certain February has the same number of days.

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How look for the general solution of a trig function