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Mathematics 17 Online
OpenStudy (anonymous):

Use an Addition or Subtraction Formula to find the exact value of the expression, cos(17pi/12)

OpenStudy (anonymous):

\[\cos \frac{ 17\pi }{ 12 }=-\cos \frac{ 5\pi }{ 12 }=-\cos (\pi/? + \pi/?)\]

OpenStudy (anonymous):

im not sure what to put in for ?'s

OpenStudy (anonymous):

What are you in?

OpenStudy (anonymous):

math 147 pre cal

OpenStudy (anonymous):

Yikes! I have no clue then....

OpenStudy (anonymous):

but i have no experience with trig identities, or anything past some algebra.. i took the wrong class :(

OpenStudy (anonymous):

its ok im winging it lol

OpenStudy (anonymous):

Whatttt that sucks! Im only in geometry and its still hard. Haha alright sounds good lol

OpenStudy (shamim):

i wanna try to solve this problem

OpenStudy (anonymous):

i reopened it if you wanna help me with it i got to a point and couldnt find the common denom

OpenStudy (shamim):

ok i m trying

OpenStudy (anonymous):

=0. just use calculator

OpenStudy (anonymous):

lol k

OpenStudy (shamim):

should i continue to do ur math

OpenStudy (anonymous):

sec

OpenStudy (anonymous):

working...

OpenStudy (shamim):

sorry not sin it will b cos

OpenStudy (anonymous):

let me work it out i think it needs to be simplified more

OpenStudy (anonymous):

cos(17pi/12) is the opening post so idk

OpenStudy (shamim):

ya i know

OpenStudy (shamim):

\[\cos(\frac{ 17\pi }{ 12 })=\cos(\frac{ 18\pi-\pi }{ 12 })=\cos(\frac{ 18\pi }{ 12 }-\frac{ \pi }{ 12})\]

OpenStudy (shamim):

\[=\cos(\frac{ 3\pi }{ 2 }-\frac{ \pi }{ 12 })\]

OpenStudy (shamim):

\[=-\sin \frac{ \pi }{ 12 }\]

OpenStudy (anonymous):

\[\cos \frac{ 17\pi }{ 12 } = -\cos \frac{ 5\pi }{ 12 } = -\cos (\frac{ \pi }{ ? } + \frac{ \pi }{ ? })\]

OpenStudy (shamim):

\[-\sin(\frac{ 3\pi }{ 12 }-\frac{ 2\pi }{ 12 })\]

OpenStudy (anonymous):

it will be a coordinate value on the unit circle

OpenStudy (anonymous):

i dont have a value for pi/12

OpenStudy (shamim):

ya u hv a value for -cos(5pi/12). bt its not shown in ur circle

OpenStudy (anonymous):

we havent gone that far :/

OpenStudy (shamim):

i think i will b able to solve ur problem in ur way too. can i try

OpenStudy (anonymous):

yes sir

OpenStudy (shamim):

ok

OpenStudy (anonymous):

i have an answer for sin(19pi/12) = \[-\frac{ \sqrt{6}+\sqrt{2} }{ 4 }\]

OpenStudy (shamim):

\[\cos(\frac{ 17\pi }{ 12 })=\cos(\frac{ 12\pi+5\pi }{ 12 })=\cos(\frac{ 12\pi }{ 12 }+\frac{ 5\pi }{ 12})\]

OpenStudy (anonymous):

what is the value of 5pi/12?

OpenStudy (anonymous):

cos value i mean

OpenStudy (shamim):

\[=\cos(\pi+\frac{ 5\pi }{ 12 })=-\cos(\frac{ 5\pi }{ 12 })=-\cos(\frac{ 3\pi }{ 12 }+\frac{ 2\pi }{ 12})\]

OpenStudy (shamim):

5pi/12=75 degree

OpenStudy (shamim):

i m going to calculate ur cos(5pi/12). please follow me. i think i will b successful

OpenStudy (shamim):

\[-\cos(\frac{ 3\pi }{ 12 })\cos(\frac{ 2\pi }{ 12 })+\sin(\frac{ 3\pi }{ 12 })\sin(\frac{ 2\pi }{ 12 })\]

OpenStudy (anonymous):

following

OpenStudy (shamim):

\[-\cos(\frac{ \pi }{ 4 })\cos(\frac{ \pi }{ 6 })+\sin(\frac{ \pi }{ 4 })\sin(\frac{ \pi }{ 6 })\]

OpenStudy (anonymous):

sweet i can work with that one

OpenStudy (shamim):

thanks

OpenStudy (shamim):

i m happy to work with u

OpenStudy (anonymous):

-(sqrt(2)/2) x (sqrt(3)/2) x (sqrt(2)/2 x (1/2)

OpenStudy (shamim):

ok i m doing

OpenStudy (anonymous):

i think there is a minus in the middle

OpenStudy (shamim):

\[-\frac{ 1 }{ \sqrt{2} }.\frac{ \sqrt{3} }{ 2 }+\frac{ 1 }{ \sqrt{2} }.\frac{ 1 }{ 2 }\]

OpenStudy (anonymous):

rationalize the denominators, we dont want roots in the denom :)

OpenStudy (shamim):

\[\frac{ 1-\sqrt{3} }{ 2\sqrt{2} }\]

OpenStudy (shamim):

ya u r right

OpenStudy (shamim):

\[=\frac{ \sqrt{2}-\sqrt{6} }{ 4 }\]

OpenStudy (anonymous):

correct! woo hoo thats mohammad

OpenStudy (anonymous):

thanks**

OpenStudy (shamim):

u r most welcome

OpenStudy (anonymous):

kinda tricky, thanks for your time and work sir

OpenStudy (shamim):

anyway may i know the name of ur country. its just my curiosity

OpenStudy (anonymous):

im in the united states

OpenStudy (shamim):

ok

OpenStudy (anonymous):

and you?

OpenStudy (shamim):

bangladesh

OpenStudy (anonymous):

thats cool its morning there? its late here

OpenStudy (shamim):

ya

OpenStudy (shamim):

its 3.20 pm here

OpenStudy (anonymous):

2.20 am here :)

OpenStudy (shamim):

lol

OpenStudy (anonymous):

just been grinding this math out lol

OpenStudy (shamim):

i enjoyed ur math too

OpenStudy (anonymous):

are you in school?

OpenStudy (shamim):

no

OpenStudy (shamim):

i m a lecturer of physics in national university

OpenStudy (anonymous):

wow that must be fun

OpenStudy (shamim):

may b

OpenStudy (anonymous):

which languages do you speak?

OpenStudy (shamim):

bangla is my mother language n english is my foreign language. so i m not good in english

OpenStudy (anonymous):

its ok i hear english is hard, i speak spanish as my foreign language

OpenStudy (shamim):

i c

OpenStudy (anonymous):

do u want to try another question

OpenStudy (shamim):

ya i wanna solve another 1

OpenStudy (shamim):

wanna try

OpenStudy (shamim):

anyway which class u r reading

OpenStudy (anonymous):

Use an Addition or Subtraction Formula to find the exact value of the expression, 165 degrees

OpenStudy (anonymous):

im in math 147 precalculus it is a hybrid class

OpenStudy (anonymous):

studying analytic trigonometry now

OpenStudy (shamim):

i think u will say sin or cos or tan or cot etc.

OpenStudy (shamim):

ok

OpenStudy (shamim):

ur class is unknown to me

OpenStudy (anonymous):

OpenStudy (anonymous):

that picture is the work of a different, but similar problem

OpenStudy (shamim):

question is not clear to me

OpenStudy (shamim):

is it sin 165

OpenStudy (shamim):

or its cos 165

OpenStudy (anonymous):

hmm

OpenStudy (anonymous):

oh cos im sorry

OpenStudy (shamim):

its ok

OpenStudy (shamim):

no problem

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