# 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 $\text{cos}\left(t\right)=-\frac{2\sqrt{10}}{7}$

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
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations