7.3 Unit circle  (Page 3/11)

 Page 3 / 11

If $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)=\frac{24}{25}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is in the fourth quadrant, find $\text{\hspace{0.17em}}\text{sin}\left(t\right).$

$\mathrm{sin}\left(t\right)=-\frac{7}{25}$

Finding sines and cosines of special angles

We have already learned some properties of the special angles, such as the conversion from radians to degrees, and we found their sines and cosines using right triangles. We can also calculate sines and cosines of the special angles using the Pythagorean Identity.

Finding sines and cosines of $\text{\hspace{0.17em}}45°\text{\hspace{0.17em}}$ Angles

First, we will look at angles of $\text{\hspace{0.17em}}45°\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\frac{\pi }{4},$ as shown in [link] . A $\text{\hspace{0.17em}}45°–45°–90°\text{\hspace{0.17em}}$ triangle is an isosceles triangle, so the x- and y -coordinates of the corresponding point on the circle are the same. Because the x- and y -values are the same, the sine and cosine values will also be equal.

At $\text{\hspace{0.17em}}t=\frac{\pi }{4},$ which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle. This means the radius lies along the line $\text{\hspace{0.17em}}y=x.\text{\hspace{0.17em}}$ A unit circle has a radius equal to 1 so the right triangle formed below the line $\text{\hspace{0.17em}}y=x\text{\hspace{0.17em}}$ has sides $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and and radius = 1. See [link] .

From the Pythagorean Theorem we get

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

We can then substitute $\text{\hspace{0.17em}}y=x.$

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

Next we combine like terms.

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

And solving for $\text{\hspace{0.17em}}x,$ we get

$\begin{array}{ccc}\hfill {x}^{2}& =& \frac{1}{2}\hfill \\ \hfill x& =& ±\frac{1}{\sqrt{2}}\hfill \end{array}$

In quadrant I, $\text{\hspace{0.17em}}x=\frac{1}{\sqrt{2}}.$

At $\text{\hspace{0.17em}}t=\frac{\pi }{4}\text{\hspace{0.17em}}$ or 45 degrees,

If we then rationalize the denominators, we get

Therefore, the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates of a point on a circle of radius $\text{\hspace{0.17em}}1\text{\hspace{0.17em}}$ at an angle of $\text{\hspace{0.17em}}45°\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).$

Finding sines and cosines of $\text{\hspace{0.17em}}30°\text{\hspace{0.17em}}$ And $\text{\hspace{0.17em}}60°\text{\hspace{0.17em}}$ Angles

Next, we will find the cosine and sine at an angle of $\text{\hspace{0.17em}}30°,$ or $\text{\hspace{0.17em}}\frac{\pi }{6}.\text{\hspace{0.17em}}$ First, we will draw a triangle inside a circle with one side at an angle of $\text{\hspace{0.17em}}30°,$ and another at an angle of $\text{\hspace{0.17em}}-30°,$ as shown in [link] . If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be $\text{\hspace{0.17em}}60°,$ as shown in [link] .

Because all the angles are equal, the sides are also equal. The vertical line has length $\text{\hspace{0.17em}}2y,$ and since the sides are all equal, we can also conclude that $\text{\hspace{0.17em}}r=2y\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=\frac{1}{2}r.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t=y,$

$\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}r$

And since $\text{\hspace{0.17em}}r=1\text{\hspace{0.17em}}$ in our unit circle,

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

Using the Pythagorean Identity, we can find the cosine value.

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 $\text{\hspace{0.17em}}30°\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right).\text{\hspace{0.17em}}$ At the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, $\text{\hspace{0.17em}}BAD,$ as shown in [link] . Angle $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ has measure $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ At point $\text{\hspace{0.17em}}B,$ we draw an angle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ with measure of $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ We know the angles in a triangle sum to $\text{\hspace{0.17em}}180°,$ so the measure of angle $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ is also $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ Now we have an equilateral triangle. Because each side of the equilateral triangle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.

The measure of angle $\text{\hspace{0.17em}}ABD\text{\hspace{0.17em}}$ is 30°. Angle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ is double angle $\text{\hspace{0.17em}}ABD,$ so its measure is 60°. $\text{\hspace{0.17em}}BD\text{\hspace{0.17em}}$ is the perpendicular bisector of $\text{\hspace{0.17em}}AC,$ so it cuts $\text{\hspace{0.17em}}AC\text{\hspace{0.17em}}$ in half. This means that $\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ the radius, or $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$ Notice that $\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$ is the x -coordinate of point $\text{\hspace{0.17em}}B,$ which is at the intersection of the 60° angle and the unit circle. This gives us a triangle $\text{\hspace{0.17em}}BAD\text{\hspace{0.17em}}$ with hypotenuse of 1 and side $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ of length $\text{\hspace{0.17em}}\frac{1}{2}.$

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
how to solve x²=2x+8 factorization?
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
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
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
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
ouky thanks braa
Rubben
I think he drive me safe
Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin