# 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 $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$

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