# 7.3 Unit circle  (Page 4/11)

 Page 4 / 11

From the Pythagorean Theorem, we get

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

Substituting $\text{\hspace{0.17em}}x=\frac{1}{2},$ we get

${\left(\frac{1}{2}\right)}^{2}+{y}^{2}=1$

Solving for $\text{\hspace{0.17em}}y,$ we get

$\begin{array}{ccc}\hfill \frac{1}{4}+{y}^{2}& =& 1\hfill \\ \hfill {y}^{2}& =& 1-\frac{1}{4}\hfill \\ \hfill {y}^{2}& =& \frac{3}{4}\hfill \\ \hfill y& =& ±\frac{\sqrt{3}}{2}\hfill \end{array}$

Since $\text{\hspace{0.17em}}t=\frac{\pi }{3}\text{\hspace{0.17em}}$ has the terminal side in quadrant I where the y- coordinate is positive, we choose $\text{\hspace{0.17em}}y=\frac{\sqrt{3}}{2},$ the positive value.

At $\text{\hspace{0.17em}}t=\frac{\pi }{3}\text{\hspace{0.17em}}$ (60°), 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 $\text{\hspace{0.17em}}60°\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right),$ so we can find the sine and cosine.

We have now found the cosine and sine values for all of the most commonly encountered angles in the first quadrant of the unit circle. [link] summarizes these values.

 Angle $0$ $\frac{\pi }{6},$ or $\text{\hspace{0.17em}}30°$ $\frac{\pi }{4},$ or $\text{\hspace{0.17em}}45°$ $\frac{\pi }{3},$ or $\text{\hspace{0.17em}}60°$ $\frac{\pi }{2},$ or $\text{\hspace{0.17em}}90°$ 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

[link] shows the common angles in the first quadrant of the unit circle.

## Using a calculator to find sine and cosine

To find the cosine and sine of angles other than the special angles, we turn to a computer or calculator. Be aware : Most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When we evaluate $\text{\hspace{0.17em}}\mathrm{cos}\left(30\right)\text{\hspace{0.17em}}$ on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.

Given an angle in radians, use a graphing calculator to find the cosine.

1. If the calculator has degree mode and radian mode, set it to radian mode.
2. Press the COS key.
3. Enter the radian value of the angle and press the close-parentheses key ")".
4. Press ENTER.

## Using a graphing calculator to find sine and cosine

Evaluate $\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{5\pi }{3}\right)\text{\hspace{0.17em}}$ using a graphing calculator or computer.

Enter the following keystrokes:

$\mathrm{cos}\left(\frac{5\pi }{3}\right)=0.5$

Evaluate $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{3}\right).$

approximately 0.866025403

## Identifying the domain and range of sine and cosine functions

Now that we can find the sine and cosine of an angle, we need to discuss their domains and ranges. What are the domains of the sine and cosine functions? That is, what are the smallest and largest numbers that can be inputs of the functions? Because angles smaller than $\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ and angles larger than $\text{\hspace{0.17em}}2\pi \text{\hspace{0.17em}}$ can still be graphed on the unit circle and have real values of $\text{\hspace{0.17em}}x,y,\text{and}\text{\hspace{0.17em}}r,$ there is no lower or upper limit to the angles that can be inputs to the sine and cosine functions. The input to the sine and cosine functions is the rotation from the positive x -axis, and that may be any real number.

What are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the unit circle, as shown in [link] . The bounds of the x -coordinate are $\text{\hspace{0.17em}}\left[-1,1\right].\text{\hspace{0.17em}}$ The bounds of the y -coordinate are also $\text{\hspace{0.17em}}\left[-1,1\right].\text{\hspace{0.17em}}$ Therefore, the range of both the sine and cosine functions is $\text{\hspace{0.17em}}\left[-1,1\right].$

## Finding reference angles

We have discussed finding the sine and cosine for angles in the first quadrant, but what if our angle is in another quadrant? For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Because the sine value is the y -coordinate on the unit circle, the other angle with the same sine will share the same y -value, but have the opposite x -value. Therefore, its cosine value will be the opposite of the first angle’s cosine value.

#### Questions & Answers

sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
Umesh Reply
I want to know trigonometry but I can't understand it anyone who can help
Siyabonga Reply
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
Sudip Reply
f(x)= 1 x    f(x)=1x  is shifted down 4 units and to the right 3 units.
Sebit Reply
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
Marty Reply
I want to know partial fraction Decomposition.
Adama Reply
classes of function in mathematics
Yazidu Reply
divide y2_8y2+5y2/y2
Sumanth Reply
wish i knew calculus to understand what's going on 🙂
Dashawn Reply
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn
what's the derivative of 4x^6
Axmed Reply
24x^5
James
10x
Axmed
24X^5
Taieb
Thanks for this helpfull app
Axmed Reply
secA+tanA=2√5,sinA=?
richa Reply
tan2a+tan2a=√3
Rahulkumar
classes of function
Yazidu
if sinx°=sin@, then @ is - ?
NAVJIT Reply
the value of tan15°•tan20°•tan70°•tan75° -
NAVJIT
0.037 than find sin and tan?
Jon Reply
cos24/25 then find sin and tan
Deepak Reply

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