# 5.2 Unit circle: sine and cosine functions  (Page 2/12)

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## Sine and cosine functions

If $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is a real number and a point $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ on the unit circle corresponds to an angle of $\text{\hspace{0.17em}}t,$ then

$\mathrm{cos}\text{\hspace{0.17em}}t=x$
$\mathrm{sin}\text{\hspace{0.17em}}t=y$

Given a point P $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ on the unit circle corresponding to an angle of $\text{\hspace{0.17em}}t,$ find the sine and cosine.

1. The sine of $t$ is equal to the y -coordinate of point $P:\mathrm{sin}\text{\hspace{0.17em}}t=y.$
2. The cosine of $t$ is equal to the x -coordinate of point

## Finding function values for sine and cosine

Point $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ is a point on the unit circle corresponding to an angle of $\text{\hspace{0.17em}}t,$ as shown in [link] . Find $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{sin}\left(t\right).\text{\hspace{0.17em}}$

We know that $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the x -coordinate of the corresponding point on the unit circle and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the y -coordinate of the corresponding point on the unit circle. So:

$\begin{array}{l}\begin{array}{l}\\ x=\mathrm{cos}\text{\hspace{0.17em}}t=\frac{1}{2}\end{array}\hfill \\ y=\mathrm{sin}\text{\hspace{0.17em}}t=\frac{\sqrt{3}}{2}\hfill \end{array}$

A certain angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ corresponds to a point on the unit circle at $\text{\hspace{0.17em}}\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\text{\hspace{0.17em}}$ as shown in [link] . Find $\mathrm{cos}\text{\hspace{0.17em}}t$ and $\mathrm{sin}\text{\hspace{0.17em}}t.$

$\mathrm{cos}\left(t\right)=-\frac{\sqrt{2}}{2},\mathrm{sin}\left(t\right)=\frac{\sqrt{2}}{2}$

## Finding sines and cosines of angles on an axis

For quadrantral angles, the corresponding point on the unit circle falls on the x- or y -axis. In that case, we can easily calculate cosine and sine from the values of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$

## Calculating sines and cosines along an axis

Find $\text{\hspace{0.17em}}\mathrm{cos}\left(90°\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{sin}\left(90°\right).\text{\hspace{0.17em}}$

Moving $\text{\hspace{0.17em}}90°\text{\hspace{0.17em}}$ counterclockwise around the unit circle from the positive x -axis brings us to the top of the circle, where the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates are (0, 1), as shown in [link] .

Using our definitions of cosine and sine,

$\begin{array}{l}x=\mathrm{cos}\text{\hspace{0.17em}}t=\mathrm{cos}\left(90°\right)=0\\ y=\mathrm{sin}\text{\hspace{0.17em}}t=\mathrm{sin}\left(90°\right)=1\end{array}$

The cosine of 90° is 0; the sine of 90° is 1.

Find cosine and sine of the angle $\text{\hspace{0.17em}}\pi .\text{\hspace{0.17em}}$

$\text{\hspace{0.17em}}\mathrm{cos}\left(\pi \right)=-1,$ $\mathrm{sin}\left(\pi \right)=0\text{\hspace{0.17em}}$

## The pythagorean identity

Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is $\text{\hspace{0.17em}}{x}^{2}+{y}^{2}=1.\text{\hspace{0.17em}}$ Because $\text{\hspace{0.17em}}x=\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}t,$ we can substitute for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ to get $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1.\text{\hspace{0.17em}}$ This equation, $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1,$ is known as the Pythagorean Identity . See [link] .

We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.

## Pythagorean identity

The Pythagorean Identity    states that, for any real number $\text{\hspace{0.17em}}t,$

${\mathrm{cos}}^{2}\text{\hspace{0.17em}}t+{\mathrm{sin}}^{2}\text{\hspace{0.17em}}t=1\text{\hspace{0.17em}}$

Given the sine of some angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and its quadrant location, find the cosine of $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$

1. Substitute the known value of $\text{\hspace{0.17em}}\mathrm{sin}\left(t\right)\text{\hspace{0.17em}}$ into the Pythagorean Identity.
2. Solve for $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right).\text{\hspace{0.17em}}$
3. Choose the solution with the appropriate sign for the x -values in the quadrant where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is located.

## Finding a cosine from a sine or a sine from a cosine

If $\text{\hspace{0.17em}}\mathrm{sin}\left(t\right)=\frac{3}{7}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is in the second quadrant, find $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right).\text{\hspace{0.17em}}$

If we drop a vertical line from the point on the unit circle corresponding to $\text{\hspace{0.17em}}t,$ we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See [link] .

Substituting the known value for sine into the Pythagorean Identity,

$\begin{array}{l}{\mathrm{cos}}^{2}\left(t\right)+{\mathrm{sin}}^{2}\left(t\right)=1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\left(t\right)+\frac{9}{49}=1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\left(t\right)=\frac{40}{49}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{cos}\left(t\right)=±\sqrt{\frac{40}{49}}=±\frac{\sqrt{40}}{7}=±\frac{2\sqrt{10}}{7}\hfill \end{array}$

Because the angle is in the second quadrant, we know the x- value is a negative real number, so the cosine is also negative. So $\text{cos}\left(t\right)=-\frac{2\sqrt{10}}{7}$

I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
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Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert