# 7.3 Unit circle  (Page 6/11)

 Page 6 / 11

## Using reference angles to find sine and cosine

1. Using a reference angle, find the exact value of $\text{\hspace{0.17em}}\mathrm{cos}\left(150°\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{sin}\left(150°\right).$
2. Using the reference angle, find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{5\pi }{4}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{5\pi }{4}.$
1. $150°\text{\hspace{0.17em}}$ is located in the second quadrant. The angle it makes with the x -axis is $\text{\hspace{0.17em}}180°-150°=30°,$ so the reference angle is $\text{\hspace{0.17em}}30°.$

This tells us that $\text{\hspace{0.17em}}150°\text{\hspace{0.17em}}$ has the same sine and cosine values as $\text{\hspace{0.17em}}30°,$ except for the sign.

$\begin{array}{ccc}\mathrm{cos}\left(30°\right)=\frac{\sqrt{3}}{2}& \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{sin}\left(30°\right)=\frac{1}{2}\end{array}$

Since $\text{\hspace{0.17em}}150°\text{\hspace{0.17em}}$ is in the second quadrant, the x -coordinate of the point on the circle is negative, so the cosine value is negative. The y -coordinate is positive, so the sine value is positive.

$\begin{array}{ccc}\mathrm{cos}\left(150°\right)=\frac{\sqrt{3}}{2}& \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{sin}\left(150°\right)=\frac{1}{2}\end{array}$
2. $\frac{5\pi }{4}\text{\hspace{0.17em}}$ is in the third quadrant. Its reference angle is $\text{\hspace{0.17em}}\frac{5\pi }{4}-\pi =\frac{\pi }{4}.\text{\hspace{0.17em}}$ The cosine and sine of $\text{\hspace{0.17em}}\frac{\pi }{4}\text{\hspace{0.17em}}$ are both $\text{\hspace{0.17em}}\frac{\sqrt{2}}{2}.\text{\hspace{0.17em}}$ In the third quadrant, both $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are negative, so:
$\begin{array}{ccc}\mathrm{cos}\phantom{\rule{0.03em}{0ex}}\frac{5\pi }{4}=-\frac{\sqrt{2}}{2}& \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{sin}\phantom{\rule{0.03em}{0ex}}\frac{5\pi }{4}=-\frac{\sqrt{2}}{2}\end{array}$
1. Use the reference angle of $\text{\hspace{0.17em}}315°\text{\hspace{0.17em}}$ to find $\text{\hspace{0.17em}}\mathrm{cos}\left(315°\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\left(315°\right).$
2. Use the reference angle of $\text{\hspace{0.17em}}-\frac{\pi }{6}\text{\hspace{0.17em}}$ to find $\text{\hspace{0.17em}}\mathrm{cos}\left(-\frac{\pi }{6}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\left(-\frac{\pi }{6}\right).$
1. $\text{cos}\left(-\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2},\mathrm{sin}\left(-\frac{\pi }{6}\right)=-\frac{1}{2}$

## Using reference angles to find coordinates

Now that we have learned how to find the cosine and sine values for special angles in the first quadrant, we can use symmetry and reference angles to fill in cosine and sine values for the rest of the special angles on the unit circle    . They are shown in [link] . Take time to learn the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates of all of the major angles in the first quadrant. Special angles and coordinates of corresponding points on the unit circle

In addition to learning the values for special angles, we can use reference angles to find $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates of any point on the unit circle, using what we know of reference angles along with the identities

First we find the reference angle corresponding to the given angle. Then we take the sine and cosine values of the reference angle, and give them the signs corresponding to the y - and x -values of the quadrant.

Given the angle of a point on a circle and the radius of the circle, find the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates of the point.

1. Find the reference angle by measuring the smallest angle to the x -axis.
2. Find the cosine and sine of the reference angle.
3. Determine the appropriate signs for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ in the given quadrant.

## Using the unit circle to find coordinates

Find the coordinates of the point on the unit circle at an angle of $\text{\hspace{0.17em}}\frac{7\pi }{6}.$

We know that the angle $\text{\hspace{0.17em}}\frac{7\pi }{6}\text{\hspace{0.17em}}$ is in the third quadrant.

First, let’s find the reference angle by measuring the angle to the x -axis. To find the reference angle of an angle whose terminal side is in quadrant III, we find the difference of the angle and $\text{\hspace{0.17em}}\pi .$

$\frac{7\pi }{6}-\pi =\frac{\pi }{6}$

Next, we will find the cosine and sine of the reference angle.

$\begin{array}{cc}\mathrm{cos}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\hfill & \phantom{\rule{1em}{0ex}}\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}\hfill \end{array}$

We must determine the appropriate signs for x and y in the given quadrant. Because our original angle is in the third quadrant, where both $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are negative, both cosine and sine are negative.

$\begin{array}{ccc}\hfill \mathrm{cos}\left(\frac{7\pi }{6}\right)& =& -\frac{\sqrt{3}}{2}\hfill \\ \hfill \mathrm{sin}\left(7\pi \right)& =& -\frac{1}{2}\hfill \end{array}$

Now we can calculate the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates using the identities $\text{\hspace{0.17em}}x=\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}\theta .$

The coordinates of the point are $\text{\hspace{0.17em}}\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\text{\hspace{0.17em}}$ on the unit circle.

Find the coordinates of the point on the unit circle at an angle of $\text{\hspace{0.17em}}\frac{5\pi }{3}.$

$\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

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