# 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

exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
1KI POWER 1/3 PLEASE SOLUTIONS
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Amit
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Dorbor
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Biswajit
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Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
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hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
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Modress
-x=7
Modress
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siame