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Posted by **JessieLynne** on Monday, January 3, 2011 at 10:42am.

What is the exact value of cos 60 as found on the unit circle?

and the value of cos 5pi/3

Please show me the steps so I understand

Thank you

- Algebra II-Reiny-Could you please show me how to do these -
**JessieLynne**, Monday, January 3, 2011 at 11:13amPlease help me with above post-thank you

- Algebra II -
**Reiny**, Monday, January 3, 2011 at 11:14amGoogle shows you many examples of the unit circle, here is one of them

http://67pics.com/view2.php?q=Picture%20Of%20The%20Unit%20Circle&url=http://www.math.tamu.edu/~austin/unit_circle.png

notice the ordered pair for 60° is (1/2 , √3/2)

your definition of cosØ = x/r

so cos 60° = (1/2) / 1 = 1/2

find 5π/3 and do the same thing

- Algebra II-Reiny could you relook at my comments,please? -
**JessieLynne**, Monday, January 3, 2011 at 11:26amI'm not suppose to copy the coordinates off the unit circle-I'm suppose to actually solve for the coordinates-like how do you get the coordinates for 60 degrees? that's where I'm confused

- Algebra II -
**Reiny**, Monday, January 3, 2011 at 11:44ammake yourself familar with the ratios of sides of the 30-60-90° and the 45-45-90° right-angled triangles

for the 30-60-90 angles the corresponding sides are

1 √3 and 2 , (notice that 1^2 + (√3)^2 = 2^2 )

so cos 60 = adjacent/hypotenuse = 1/2

sin 60 = opposite/hypotenuse = √3/2

tan 60 = opp/adj = √3/1 = √3

you can do the same thing for 30 and 45, and get all those special angles in the first quadrants.

Once you get into the other quadrants, the only thing that will change are the signs of the numbers.

- Algebra II -
**Reiny**, Monday, January 3, 2011 at 11:53amthese two pages might help for the triangles I described

http://www.themathpage.com/atrig/30-60-90-triangle.htm

http://www.themathpage.com/atrig/isosceles-right-triangle.htm

- Algebra II -
**mathematical mom**, Monday, January 3, 2011 at 1:27pmdraw a circle with a raduis of one( from ((0,0) go left,right,up,down, one unit.

draw a triangle by connecting points (0,0)(1,1) and (1,0). you now have a right triangle with a hypotenuse of 1

(also radius on circle) the short side

(1,1)to(1,0) is 1/2.(Geometry TH short

leg is 1/2 hypotenuse)long leg is sq root of 3, from (0,0) to (1,0),which is

sq root of 3 times short leg.(Geometry TH).30 degrees is near (0,0) because is it opposite the shortest side 1/2. 60 degrees is near (1,1) and the hypotenuse is opp the right angle. Now the cosine is defined as adjacent over the hypotenuse. so the

adjacent angle (closet to 60, not opp or

the hyp) is 1/2 and the hyp is 1, so cosine of 60 degress is 1/2 over 1 which is 1/2. If you draw the picture it should make sense. hope this works.