# 7.3 Unit circle  (Page 7/11)

 Page 7 / 11

Access these online resources for additional instruction and practice with sine and cosine functions.

## Key equations

 Cosine $\mathrm{cos}\text{\hspace{0.17em}}t=x$ Sine $\mathrm{sin}\text{\hspace{0.17em}}t=y$ Pythagorean Identity ${\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1$

## Key concepts

• Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit.
• Using the unit circle, the sine of an angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ equals the y -value of the endpoint on the unit circle of an arc of length $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ whereas the cosine of an angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ equals the x -value of the endpoint. See [link] .
• The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. See [link] .
• When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. See [link] .
• Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See [link] .
• The domain of the sine and cosine functions is all real numbers.
• The range of both the sine and cosine functions is $\text{\hspace{0.17em}}\left[-1,1\right].$
• The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.
• The signs of the sine and cosine are determined from the x - and y -values in the quadrant of the original angle.
• An angle’s reference angle is the size angle, $\text{\hspace{0.17em}}t,$ formed by the terminal side of the angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and the horizontal axis. See [link] .
• Reference angles can be used to find the sine and cosine of the original angle. See [link] .
• Reference angles can also be used to find the coordinates of a point on a circle. See [link] .

## Verbal

Describe the unit circle.

The unit circle is a circle of radius 1 centered at the origin.

What do the x- and y- coordinates of the points on the unit circle represent?

Discuss the difference between a coterminal angle and a reference angle.

Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, $\text{\hspace{0.17em}}t,$ formed by the terminal side of the angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and the horizontal axis.

Explain how the cosine of an angle in the second quadrant differs from the cosine of its reference angle in the unit circle.

Explain how the sine of an angle in the second quadrant differs from the sine of its reference angle in the unit circle.

The sine values are equal.

## Algebraic

For the following exercises, use the given sign of the sine and cosine functions to find the quadrant in which the terminal point determined by $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ lies.

$\text{sin}\left(t\right)<0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{cos}\left(t\right)<0$

$\text{sin}\left(t\right)>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)>0$

I

$\text{sin}\left(t\right)>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)<0$

$\text{sin}\left(t\right)>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)>0$

IV

For the following exercises, find the exact value of each trigonometric function.

$\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{2}$

$\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{3}$

$\frac{\sqrt{3}}{2}$

$\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{2}$

$\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{3}$

$\frac{1}{2}$

$\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{4}$

$\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{4}$

$\frac{\sqrt{2}}{2}$

$\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{6}$

$\mathrm{sin}\text{\hspace{0.17em}}\pi$

0

$\mathrm{sin}\text{\hspace{0.17em}}\frac{3\pi }{2}$

$\mathrm{cos}\text{\hspace{0.17em}}\pi$

-1

$\mathrm{cos}\text{\hspace{0.17em}}0$

$\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{6}$

$\frac{\sqrt{3}}{2}$

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
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
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
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
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
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1