# 5.2 Unit circle: sine and cosine functions  (Page 3/12)

 Page 3 / 12

If $\mathrm{cos}\left(t\right)=\frac{24}{25}$ and $t$ is in the fourth quadrant, find $\text{sin}\left(t\right).$

$\mathrm{sin}\left(t\right)=-\frac{7}{25}$

## Finding sines and cosines of special angles

We have already learned some properties of the special angles, such as the conversion from radians to degrees. We can also calculate sines and cosines of the special angles using the Pythagorean Identity    and our knowledge of triangles.

## Finding sines and cosines of 45° angles

First, we will look at angles of $\text{\hspace{0.17em}}45°\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\frac{\pi }{4},$ as shown in [link] . A $\text{\hspace{0.17em}}45°–45°–90°\text{\hspace{0.17em}}$ triangle is an isosceles triangle, so the x- and y -coordinates of the corresponding point on the circle are the same. Because the x- and y -values are the same, the sine and cosine values will also be equal.

At $\text{\hspace{0.17em}}t=\frac{\pi }{4}$ , which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle    . This means the radius lies along the line $\text{\hspace{0.17em}}y=x.\text{\hspace{0.17em}}$ A unit circle has a radius equal to 1. So, the right triangle formed below the line $\text{\hspace{0.17em}}y=x\text{\hspace{0.17em}}$ has sides $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and and a radius = 1. See [link] .

From the Pythagorean Theorem we get

${x}^{2}+{y}^{2}=1$

Substituting $\text{\hspace{0.17em}}y=x,$ we get

${x}^{2}+{x}^{2}=1$

Combining like terms we get

$2{x}^{2}=1$

And solving for $\text{\hspace{0.17em}}x,$ we get

In quadrant I, $\text{\hspace{0.17em}}x=\frac{1}{\sqrt{2}}.\text{\hspace{0.17em}}$

At $\text{\hspace{0.17em}}t=\frac{\pi }{4}\text{\hspace{0.17em}}$ or 45 degrees,

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

If we then rationalize the denominators, we get

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

Therefore, the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates of a point on a circle of radius $\text{\hspace{0.17em}}1\text{\hspace{0.17em}}$ at an angle of $\text{\hspace{0.17em}}45°\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).\text{\hspace{0.17em}}$

## Finding sines and cosines of 30° and 60° angles

Next, we will find the cosine and sine at an angle of $\text{\hspace{0.17em}}30°,$ or $\text{\hspace{0.17em}}\frac{\pi }{6}\text{\hspace{0.17em}}$ . First, we will draw a triangle inside a circle with one side at an angle of $\text{\hspace{0.17em}}30°,$ and another at an angle of $\text{\hspace{0.17em}}-30°,$ as shown in [link] . If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be $\text{\hspace{0.17em}}60°,$ as shown in [link] .

Because all the angles are equal, the sides are also equal. The vertical line has length $\text{\hspace{0.17em}}2y,$ and since the sides are all equal, we can also conclude that $\text{\hspace{0.17em}}r=2y\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=\frac{1}{2}r.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t=y\text{\hspace{0.17em}}$ ,

$\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}r$

And since $\text{\hspace{0.17em}}r=1\text{\hspace{0.17em}}$ in our unit circle    ,

Using the Pythagorean Identity, we can find the cosine value.

The $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates for the point on a circle of radius $\text{\hspace{0.17em}}1\text{\hspace{0.17em}}$ at an angle of $\text{\hspace{0.17em}}30°\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right).\text{\hspace{0.17em}}$ At $t=\frac{\pi }{3}$ (60°), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, $\text{\hspace{0.17em}}BAD,$ as shown in [link] . Angle $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ has measure $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ At point $\text{\hspace{0.17em}}B,$ we draw an angle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ with measure of $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ We know the angles in a triangle sum to $\text{\hspace{0.17em}}180°,$ so the measure of angle $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ is also $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ Now we have an equilateral triangle. Because each side of the equilateral triangle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.

The measure of angle $\text{\hspace{0.17em}}ABD\text{\hspace{0.17em}}$ is 30°. So, if double, angle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ is 60°. $\text{\hspace{0.17em}}BD\text{\hspace{0.17em}}$ is the perpendicular bisector of $\text{\hspace{0.17em}}AC,$ so it cuts $\text{\hspace{0.17em}}AC\text{\hspace{0.17em}}$ in half. This means that $\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ the radius, or $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$ Notice that $\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$ is the x -coordinate of point $\text{\hspace{0.17em}}B,$ which is at the intersection of the 60° angle and the unit circle. This gives us a triangle $\text{\hspace{0.17em}}BAD\text{\hspace{0.17em}}$ with hypotenuse of 1 and side $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ of length $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$

#### Questions & Answers

The average annual population increase of a pack of wolves is 25.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
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
Jhon Reply
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
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
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?
Diddy Reply
6000
Robert
more than 6000
Robert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
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.
rachel Reply
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?
Reena Reply
what is foci?
Reena Reply
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
Bryssen Reply

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