# 7.3 Unit circle  (Page 2/11)

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## Sine and cosine functions

If $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is a real number and a point $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ on the unit circle corresponds to a central angle $\text{\hspace{0.17em}}t,$ then

$\mathrm{cos}\text{\hspace{0.17em}}t=x$
$\mathrm{sin}\text{\hspace{0.17em}}t=y$

Given a point P $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ on the unit circle corresponding to an angle of $\text{\hspace{0.17em}}t,$ find the sine and cosine.

1. The sine of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is equal to the y -coordinate of point
2. The cosine of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is equal to the x -coordinate of point

## Finding function values for sine and cosine

Point $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ is a point on the unit circle corresponding to an angle of $\text{\hspace{0.17em}}t,$ as shown in [link] . Find $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{sin}\left(t\right).$

We know that $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the x -coordinate of the corresponding point on the unit circle and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the y -coordinate of the corresponding point on the unit circle. So:

$\begin{array}{ccc}\hfill x& =\mathrm{cos}\text{\hspace{0.17em}}t\hfill & =\frac{1}{2}\hfill \\ y& =\mathrm{sin}\text{\hspace{0.17em}}t\hfill & =\frac{\sqrt{3}}{2}\hfill \end{array}$

A certain angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ corresponds to a point on the unit circle at $\text{\hspace{0.17em}}\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\text{\hspace{0.17em}}$ as shown in [link] . Find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t.$

$\mathrm{cos}\left(t\right)=-\frac{\sqrt{2}}{2},\mathrm{sin}\left(t\right)=\frac{\sqrt{2}}{2}$

## Finding sines and cosines of angles on an axis

For quadrantral angles, the corresponding point on the unit circle falls on the x- or y -axis. In that case, we can easily calculate cosine and sine from the values of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.$

## Calculating sines and cosines along an axis

Find $\text{\hspace{0.17em}}\text{cos}\left(90°\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{sin}\left(90°\right).$

Moving $\text{\hspace{0.17em}}90°\text{\hspace{0.17em}}$ counterclockwise around the unit circle from the positive x -axis brings us to the top of the circle, where the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates are $\text{\hspace{0.17em}}\left(0,1\right),$ as shown in [link] .

We can then use our definitions of cosine and sine.

The cosine of $\text{\hspace{0.17em}}90°\text{\hspace{0.17em}}$ is 0; the sine of $\text{\hspace{0.17em}}90°\text{\hspace{0.17em}}$ is 1.

Find cosine and sine of the angle $\text{\hspace{0.17em}}\pi .$

$\mathrm{cos}\left(\pi \right)=-1,\mathrm{sin}\left(\pi \right)=0$

## The pythagorean identity

Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is $\text{\hspace{0.17em}}{x}^{2}+{y}^{2}=1.\text{\hspace{0.17em}}$ Because $\text{\hspace{0.17em}}x=\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}t,$ we can substitute for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ to get $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1.\text{\hspace{0.17em}}$ This equation, $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1,$ is known as the Pythagorean Identity    . See [link] .

We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.

## Pythagorean identity

The Pythagorean Identity    states that, for any real number $\text{\hspace{0.17em}}t,$

${\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1$

Given the sine of some angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and its quadrant location, find the cosine of $\text{\hspace{0.17em}}t.$

1. Substitute the known value of $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ into the Pythagorean Identity.
2. Solve for $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t.$
3. Choose the solution with the appropriate sign for the x -values in the quadrant where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is located.

## Finding a cosine from a sine or a sine from a cosine

If $\text{\hspace{0.17em}}\mathrm{sin}\left(t\right)=\frac{3}{7}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is in the second quadrant, find $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right).$

If we drop a vertical line from the point on the unit circle corresponding to $\text{\hspace{0.17em}}t,$ we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See [link] .

Substituting the known value for sine into the Pythagorean Identity,

$\begin{array}{ccc}\hfill {\text{cos}}^{2}\left(t\right)+{\mathrm{sin}}^{2}\left(t\right)& =& 1\hfill \\ \hfill {\text{cos}}^{2}\left(t\right)+\frac{9}{49}& =& 1\hfill \\ \hfill {\text{cos}}^{2}\left(t\right)& =& \frac{40}{49}\hfill \\ \hfill \text{cos}\left(t\right)& =& ±\sqrt{\frac{40}{49}}=±\frac{\sqrt{40}}{7}=±\frac{2\sqrt{10}}{7}\hfill \end{array}$

Because the angle is in the second quadrant, we know the x- value is a negative real number, so the cosine is also negative.

$\text{cos}\left(t\right)=-\frac{2\sqrt{10}}{7}$

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
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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
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1