Mathematics
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OpenStudy (anonymous):
In the Half Angle ID how do you know whether to use the + or - sign?
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hartnn (hartnn):
ID ?
OpenStudy (anonymous):
anyone :c
Parth (parthkohli):
identity.
Parth (parthkohli):
Both signs.
hartnn (hartnn):
half angle identity of what ?
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Parth (parthkohli):
\(\pm\) is just one addition that something applies to both signs.
OpenStudy (anonymous):
yea but youd have to chose wether its - or +
Parth (parthkohli):
You have to choose both. See, it's actually the point that a square root can be both positive and negative.
Parth (parthkohli):
For example, the square root of 64 is \(\textbf{both}\) 8 and -8. If you're lazy, then you say that it is \(\pm 8\). Get the point?
hartnn (hartnn):
where did half angle come in this picture ?
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Parth (parthkohli):
sin(x/2) and cos(x/2) @hartnn
OpenStudy (anonymous):
Okay so like if you were to solve Cos (13pi/12 )
OpenStudy (anonymous):
how do you know it simplifies to -\[-\sqrt{1+\cos (13\Pi/6)}/2\]
OpenStudy (anonymous):
Like how do you decide theres a (-)
hartnn (hartnn):
depending on the quadrant in which the angle lies.
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hartnn (hartnn):
13 pi/12 lies in which quadrant ?
OpenStudy (anonymous):
1?
OpenStudy (anonymous):
Because you have to mult by 1/2
hartnn (hartnn):
13pi/12 = pi + pi/12 , right ?
OpenStudy (anonymous):
you mean 1/2 times pi/ 13 ... Like thats how to simplify it
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OpenStudy (anonymous):
oops i mennt 1/2 times 13pi/6
hartnn (hartnn):
before doing 1/2 times..... whatever,
u need to find where your original angle lies., in which quadrant.
OpenStudy (anonymous):
oh
hartnn (hartnn):
13pi/12 = pi+pi/12
lies in which quadrant ?
hartnn (hartnn):
hint : 1st Quad : 0 to pi/2
2nd Quad : pi/2 to pi
3rd Quad : pi to 3pi/2
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OpenStudy (anonymous):
if would be Q2 but like how does 13pi/12 = pi+pi/12
hartnn (hartnn):
pi + some angle will go beyond 2nd Quadrant
so its in 3rd Quadrant, ok ?
OpenStudy (anonymous):
Oh okay
OpenStudy (anonymous):
but still dont get how you did 13pi/12 = pi+pi/12
hartnn (hartnn):
that is simple arithmetaic, since, 13pi/12 was greater than pi, i subtracted pi.
and got pi/12
so, 13pi/12 = pi+pi/12
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hartnn (hartnn):
and in 3rd Quadrant, cosine is negative...
hence, you select, - **minus**
OpenStudy (anonymous):
Lol okayyes can you give me another problem so I can practice how to find with Q its in
hartnn (hartnn):
yes,
13pi/7
which Quad ?
OpenStudy (anonymous):
wuld it also be in Q 3...
hartnn (hartnn):
nopes, how Q 3 ?
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OpenStudy (anonymous):
I thought of it like pi + pi/7
hartnn (hartnn):
can you convert radians into degrees ?
OpenStudy (anonymous):
man hhow come I dont understand how to do it x_x
OpenStudy (anonymous):
oh yes
hartnn (hartnn):
i will give a better method,
can you convert radians into degrees ?
then convert and tell me what u get.
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OpenStudy (anonymous):
Wait would it be in the Q2
hartnn (hartnn):
no.
hartnn (hartnn):
13pi/7 = 13*180/7 = 334.28...now 334 is in which Q ?
hartnn (hartnn):
its less than 360 and > 270
OpenStudy (anonymous):
the 4th ._.
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hartnn (hartnn):
yup.
so convert the angle into degrees and then its easy...
hartnn (hartnn):
one more ?
hartnn (hartnn):
7pi/6 ?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
okay so I would use the method part / whole
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OpenStudy (anonymous):
so 7pi/6 / 2pi = x / 360
hartnn (hartnn):
?
pi radians = 180 degrees
7*180/6 =210
210 in which Quadrant ?
OpenStudy (anonymous):
I got the Q3
hartnn (hartnn):
yes.
got it how to find which Q ?
OpenStudy (anonymous):
wait one more lol
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hartnn (hartnn):
ok, 3pi/4
OpenStudy (anonymous):
lol I alreadyady know that one its in Q2
hartnn (hartnn):
lol
then pi/12
OpenStudy (anonymous):
Q1
hartnn (hartnn):
yes, last
5pi/3
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OpenStudy (anonymous):
oh srry I am also doing my essay lol but um okay for thisone it will beeee
OpenStudy (anonymous):
um so it would be in Q2
hartnn (hartnn):
no, try again.
OpenStudy (anonymous):
WAIT I MEANT Q$
OpenStudy (anonymous):
*Q4
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hartnn (hartnn):
excellent!
now i award you Masters Degree in finding Quadrant of an angle. :P
you can continue with your essay....
OpenStudy (anonymous):
lol thnkx btu I have more math questions.. I hve a test today x_x
hartnn (hartnn):
\[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#e6e600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]
sure, do ask...
OpenStudy (anonymous):
okay I'll make new questions
hartnn (hartnn):
:)