# 5.3 The other trigonometric functions  (Page 2/13)

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The point $\text{\hspace{0.17em}}\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\text{\hspace{0.17em}}$ is on the unit circle, as shown in [link] . Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t.$

$\mathrm{sin}\text{\hspace{0.17em}}t=-\frac{\sqrt{2}}{2},\mathrm{cos}\text{\hspace{0.17em}}t=\frac{\sqrt{2}}{2},\mathrm{tan}\text{\hspace{0.17em}}t=-1,\mathrm{sec}\text{\hspace{0.17em}}t=\sqrt{2},\mathrm{csc}\text{\hspace{0.17em}}t=-\sqrt{2},\mathrm{cot}\text{\hspace{0.17em}}t=-1$

## Finding the trigonometric functions of an angle

Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}t=\frac{\pi }{6}.$

We have previously used the properties of equilateral triangles to demonstrate that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{6}=\frac{1}{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{6}=\frac{\sqrt{3}}{2}.\text{\hspace{0.17em}}$ We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values.

$\begin{array}{l}\mathrm{sec}\frac{\pi }{6}=\frac{1}{\mathrm{cos}\frac{\pi }{6}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\hfill \end{array}$
$\mathrm{csc}\frac{\pi }{6}=\frac{1}{\mathrm{sin}\frac{\pi }{6}}=\frac{1}{\frac{1}{2}}=2$
$\begin{array}{l}\mathrm{cot}\frac{\pi }{6}=\frac{\mathrm{cos}\frac{\pi }{6}}{\mathrm{sin}\frac{\pi }{6}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\hfill \end{array}$

Find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\mathrm{cos}\text{\hspace{0.17em}}t,\mathrm{tan}\text{\hspace{0.17em}}t,\mathrm{sec}\text{\hspace{0.17em}}t,\mathrm{csc}\text{\hspace{0.17em}}t,$ and $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}t=\frac{\pi }{3}.$

$\begin{array}{l}\mathrm{sin}\frac{\pi }{3}=\frac{\sqrt{3}}{2}\\ \mathrm{cos}\frac{\pi }{3}=\frac{1}{2}\\ \mathrm{tan}\frac{\pi }{3}=\sqrt{3}\\ \mathrm{sec}\frac{\pi }{3}=2\\ \mathrm{csc}\frac{\pi }{3}=\frac{2\sqrt{3}}{3}\\ \mathrm{cot}\frac{\pi }{3}=\frac{\sqrt{3}}{3}\end{array}$

Because we know the sine and cosine values for the common first-quadrant angles, we can find the other function values for those angles as well by setting $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ equal to the cosine and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in [link] .

 Angle $\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ Cosine 1 $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{1}{2}$ 0 Sine 0 $\frac{1}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{3}}{2}$ 1 Tangent 0 $\frac{\sqrt{3}}{3}$ 1 $\sqrt{3}$ Undefined Secant 1 $\frac{2\sqrt{3}}{3}$ $\sqrt{2}$ 2 Undefined Cosecant Undefined 2 $\sqrt{2}$ $\frac{2\sqrt{3}}{3}$ 1 Cotangent Undefined $\sqrt{3}$ 1 $\frac{\sqrt{3}}{3}$ 0

## Using reference angles to evaluate tangent, secant, cosecant, and cotangent

We can evaluate trigonometric functions of angles outside the first quadrant using reference angles as we have already done with the sine and cosine functions. The procedure is the same: Find the reference angle    formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by x - and y -values in the original quadrant. [link] shows which functions are positive in which quadrant.

To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “A Smart Trig Class.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is “ A ,” a ll of the six trigonometric functions are positive. In quadrant II, “ S mart,” only s ine and its reciprocal function, cosecant, are positive. In quadrant III, “ T rig,” only t angent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “ C lass,” only c osine and its reciprocal function, secant, are positive.

Given an angle not in the first quadrant, use reference angles to find all six trigonometric functions.

1. Measure the angle formed by the terminal side of the given angle and the horizontal axis. This is the reference angle.
2. Evaluate the function at the reference angle.
3. Observe the quadrant where the terminal side of the original angle is located. Based on the quadrant, determine whether the output is positive or negative.

## Using reference angles to find trigonometric functions

Use reference angles to find all six trigonometric functions of $\text{\hspace{0.17em}}-\frac{5\pi }{6}.\text{\hspace{0.17em}}$

The angle between this angle’s terminal side and the x -axis is $\text{\hspace{0.17em}}\frac{\pi }{6},$ so that is the reference angle. Since $\text{\hspace{0.17em}}-\frac{5\pi }{6}\text{\hspace{0.17em}}$ is in the third quadrant, where both $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are negative, cosine, sine, secant, and cosecant will be negative, while tangent and cotangent will be positive.

$\begin{array}{l}\mathrm{cos}\left(-\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2},\mathrm{sin}\left(-\frac{5\pi }{6}\right)=-\frac{1}{2},\text{tan}\left(-\frac{5\pi }{6}\right)=\frac{\sqrt{3}}{3}\hfill \\ \text{sec}\left(-\frac{5\pi }{6}\right)=-\frac{2\sqrt{3}}{3},\text{csc}\left(-\frac{5\pi }{6}\right)=-2,\mathrm{cot}\left(-\frac{5\pi }{6}\right)=\sqrt{3}\hfill \end{array}$

difference between calculus and pre calculus?
give me an example of a problem so that I can practice answering
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott