Mathematics
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OpenStudy (anonymous):
Use an Addition or Subtraction Formula to find the exact value of the expression,
cos(17pi/12)
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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....
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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
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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
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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
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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):
\[-\sin(\frac{ 3\pi }{ 12 }-\frac{ 2\pi }{ 12 })\]
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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
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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?
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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):
\[-\cos(\frac{ 3\pi }{ 12 })\cos(\frac{ 2\pi }{ 12 })+\sin(\frac{ 3\pi }{ 12 })\sin(\frac{ 2\pi }{ 12 })\]
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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
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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 :)
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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**
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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
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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
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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?
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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?
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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
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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
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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
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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
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OpenStudy (shamim):
its ok
OpenStudy (shamim):
no problem