Evaluate 2sin(pi/3)cos(pi/3)

- anonymous

Evaluate 2sin(pi/3)cos(pi/3)

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- anonymous

@jim_thompson5910 Hii are you busy? Up for any IB SL Math questions tonight?

- freckles

you could use unit circle to evaluate both sin(pi/3) and cos(pi/3)

- freckles

you could also use the double angle identity
and use unit circle once

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## More answers

- anonymous

Yes, I have it in front of me but I'm confused on what to do

- freckles

can you find pi/3 on the unit circle?

- anonymous

yea it's 60 degrees right?

- freckles

there should be an ordered pair designated to it

- freckles

(x,y)=(cos(),sin())

- anonymous

1/2, sqrt3/2

- freckles

so that means the cos(pi/3) is 1/2
and sin(pi/3) is sqrt(3)/2
plug in

- anonymous

ok gimme a sec

- anonymous

ok, just for clarification, what exactly am I doing once I plug them in? I mean am I just dividing, multiplying, etc? By the way, the answer in the back is (0.766, -0.643)

- anonymous

uhh hello..?

- freckles

2sin(pi/3)cos(pi/3)
is multiplication

- freckles

abc means a times b times c

- anonymous

Ohh ok my bad. so i got .037

- freckles

how did you get that?

- anonymous

I just plugged into my calculator 2sin(pi/3) * cos(pi/3)

- anonymous

its not right, i know ;-;

- freckles

sin(pi/3) is sqrt(3)/2 so replace sin(pi/3) with sqrt(3)/2
cos(pi/3) is 1/2 so replace cos(pi/3) with 1/2
\[2[\sin(\frac{\pi}{3})][\sin(\frac{\pi}{3})] \\ 2[ \frac{\sqrt{3}}{2}][\frac{1}{2}] = \frac{2}{2} [\sqrt{3}][\frac{1}{2}]=....\]
you can simplify a little

- freckles

oops one of those sin were suppose to read cos

- anonymous

the second one right?

- UsukiDoll

\[2[\sin(\frac{\pi}{3})][\cos(\frac{\pi}{3})] \\ 2[ \frac{\sqrt{3}}{2}][\frac{1}{2}] = \frac{2}{2} [\sqrt{3}][\frac{1}{2}]=....\]

- anonymous

ahh ok so did you get rid of the 2 in the beginning and then add in the 2/2 as a separate thing?

- freckles

multiplication is commutative ab =ba

- freckles

\[2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \\ =\frac{2}{1} \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \\ =\frac{2 }{2 } \cdot \frac{\sqrt{3}}{1} \cdot \frac{1}{2}\]

- anonymous

gotcha. But at any rate, I ended up getting 4.71

- anonymous

which, again, doesn't seem right since the answer in the back of the book is totally different

- freckles

you do know 2/2=?

- anonymous

1

- freckles

right so you have
\[\sqrt{3} \cdot \frac{1}{2} \text{ which is just } \frac{\sqrt{3}}{2}\]
I don't see how you are getting a number bigger than 1 ...

- freckles

since sqrt(3)/2 is definitely less than 1

- anonymous

i think it's just cos im plugging it into my scientific calc then

- freckles

ok

- anonymous

oh wait im so dumb i was doing pi3

- anonymous

im so sorry my bad

- anonymous

its .866 right

- freckles

yes sqrt(3)/2 is approximately .866

- UsukiDoll

yeah

- anonymous

ok but the answer says it's .766, -0.643

- jim_thompson5910

@brriiarr the fact that it shows two answers, when it should be one answer, suggests that maybe there's a mixup between the problem and answer

- freckles

I don't know what your book is talking about the answer is sqrt(3)/2 which could be written approximately as .866

- UsukiDoll

maybe it's a typo for .766 ???

- UsukiDoll

Ah yes, sometimes there are typos in book. No book is published perfectly. sigh

- anonymous

crap ok. it says in a weird little font next to it sqrt3/2 but i guess you guys were right. umm..do you think i could ask for help on one more?

- freckles

you might also want to check the question you type was the question you meant
and the answer that correspond to the question you asked was the right correspondence you chose

- anonymous

i think it really was just a typo :p

- anonymous

um but yea I have on more i needed a little help on

- anonymous

ok lol so it's just to find the perimeter and area of the sector. I'll draw it

- anonymous

rty|dw:1449550468325:dw|

- anonymous

ahh so yea, i dunno how to do this at all

- anonymous

umm are any of you guys free or should i just ask another time

- jim_thompson5910

so this is circle, well a piece of a circle
what is the radius of this circle sector? I see 4 but I also see 1^c

- anonymous

it doesnt say

- anonymous

my book says that for theta in radians, the arc length is l=theta*radius but i dunno if that helps

- jim_thompson5910

so this problem is in your book? or on a computer?
if possible, please post what the full problem looks like

- anonymous

here, ill take a pic. gimme a minute.

- anonymous

##### 1 Attachment

- jim_thompson5910

thanks

- jim_thompson5910

I'm going to guess that they meant to say \(\LARGE 1^{\circ}\) instead of \(\LARGE 1^c\)

- anonymous

yea it's an aussie book so they write things weird

- anonymous

ah man is this a long one :/

- jim_thompson5910

oh, I did not know that
I'm trying to verify that, but I'm noticing that Australian geometry books also use the degree symbol. So maybe it's only some Australian books?

- anonymous

maaybeee

- jim_thompson5910

anyways, if it is indeed 1 degree as the central angle, then you'll need to convert 1 degree to radians
1 degree = (1 degree)*(pi radians/180 degrees) = pi/180 = 0.0174532925 radians approximately

- jim_thompson5910

now that we know
1 degree = 0.0174532925 radians
we can use the formula you wrote out
L = theta*r
L = arc length
theta = central angle in radians
r = radius

- anonymous

ok one sec lemme write this stuff down

- jim_thompson5910

to convert from degrees to radians, I used the identity
pi radians = 180 degrees

- jim_thompson5910

and I'm not even thinking, they used the degree symbol just below in set 8B on the same image attachment you posted
so it's probably another typo

- anonymous

yes, i understood that. so now what do you multiply? and ugh yea my textbook is horrible honestly i hate it

- jim_thompson5910

L = theta*r
multiply the angle in radians by the radius

- anonymous

i mean, for ex, what's 4 supposed to be?

- jim_thompson5910

I have a feeling it's not really 1 degree. That just seems too small

- anonymous

wait whats the radius ;-; is that 0.01745...

- jim_thompson5910

4 looks like the radius

- anonymous

ok right that makes sense. so l=4*0.017453425?

- jim_thompson5910

yes

- anonymous

kk one sec

- anonymous

so .0698137

- anonymous

lol it says the perimeter = 12 units, area = 8units^2

- jim_thompson5910

so theta is not really 1 degree

- jim_thompson5910

maybe theta = 1 radian
if so
L = theta*r
L = 1*4
L = 4

- jim_thompson5910

curved portion = 4 units
straight portions = 2*4 = 8 units

- jim_thompson5910

I've honestly never seen "1 radian" written as \(\LARGE 1^c\) but maybe that notation is how australians write it

- anonymous

wait so i kinda get it, but why is the curved portion 4 units? i thought the 4 was the radius for the straight line..

- jim_thompson5910

hmm I learned something new
http://image.slidesharecdn.com/pmb05anglesmeasure-151011170600-lva1-app6891/95/pm-b05-angles-measurement-6-638.jpg

- anonymous

ahh cool haha now i know too ^-^

- jim_thompson5910

because
arc length = (angle in radians)*(radius)
arc length = (1)*(4)
arc length = 4

- jim_thompson5910

there are 2 sides that are straight and equal to each other
|dw:1449552118816:dw|

- anonymous

jesus christ i hate this stupid thing ok so lemme get this straight. L=\[\theta*r\] =4, correct? so then you multiply by 2 cos o f the 2 congruent sides

- anonymous

and then for perimeter, there's 3 sides, the curved, and the two straight lines
so therefore it's the arc length^3

- anonymous

so 4*3 = 12 = perimeter

- jim_thompson5910

yes, the curved side is 4 units long
each of the 2 straight sides are 4 units long
in total, 4+4+4 = 12

- jim_thompson5910

it's coincidental how the curved side is equal to the 2 other sides

- jim_thompson5910

that only happens when the angle is 1 radian

- anonymous

ok, I gotcha. anyway, thank you so much. I wish I could give you the best response :/

- jim_thompson5910

if you look at the animation on this page
https://en.wikipedia.org/wiki/Radian
it shows how the radian is set up

- anonymous

thank you! anyway i might be on and off openstudy for a bit. how late do you usually stay? it's 12:30 am eastern time here

- anonymous

im just wondering cos i dont' wanna bother you right when you're about to get to bed lol

- jim_thompson5910

I've got another hour or so, don't worry

- anonymous

kk, cool. ill probably tag ya later. bye for now!!

- jim_thompson5910

alright, good luck

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