rishabh.mission
  • rishabh.mission
matrix Question ....plz help
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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rishabh.mission
  • rishabh.mission
See the attachment
1 Attachment
rishabh.mission
  • rishabh.mission
please just solve this Question
phi
  • phi
one way is use the trig identities \[ \tan \left(\frac{x}{2} \right)= \frac{1-\cos\ x}{\sin\ x }=\frac{\sin\ x }{1+\cos\ x}\]

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rishabh.mission
  • rishabh.mission
plz buddy solve this i don't have any idea how to solve..
phi
  • phi
you could start by writing down I-A can you do that ?
rishabh.mission
  • rishabh.mission
yes i did
rishabh.mission
  • rishabh.mission
I have problem with that equation in the circle i shown.
1 Attachment
phi
  • phi
call 2 sin^2(a/2) = x then it looks like 1-x+x
phi
  • phi
**look at the top left expression. Doesn't that simplify?
rishabh.mission
  • rishabh.mission
:( not understand
phi
  • phi
or are you asking how they got to that point ?
rishabh.mission
  • rishabh.mission
yes
rishabh.mission
  • rishabh.mission
:)
rishabh.mission
  • rishabh.mission
that's Eq^n 1st
1 Attachment
phi
  • phi
you just posted the left side. on the right side (the messy one) they did I-A times the other matrix I take it you understand how they got to the matrix labeled (2) ?
phi
  • phi
the matrix above the circled stuff....
rishabh.mission
  • rishabh.mission
hmm
phi
  • phi
We could go through that mess, but I would replace tan(a/2) with (1-cos)/sin in the top left entry: cos a + sin a tan(a/2) becomes cos a + sin a * (1-cos a)/sina= cosa + 1 -cosa = 1 which is what we want. Doing the way you have laid out is messier, but we can go through that if you want
rishabh.mission
  • rishabh.mission
can we write \[\cos \alpha/2 = (1-\tan ^{2} \alpha/2) / (1+\tan^{2} \alpha/2)\]
rishabh.mission
  • rishabh.mission
sorry ** its \[\cos \alpha \]
anonymous
  • anonymous
can I say something?
rishabh.mission
  • rishabh.mission
yes please .. @hoa
anonymous
  • anonymous
left side, i got I+A = \[\left[\begin{matrix}1 & -\tan \frac{ \alpha }{ 2} \\ \tan \frac{ \alpha }{ 2 } & 1\end{matrix}\right]\] = right side = (I-A) [ ] = \[\left[\begin{matrix}1 & \tan \frac{ \alpha }{ 2}\\ -\tan \frac{ \alpha }{ 2 } & 1\end{matrix}\right]\left[\begin{matrix}\cos & -\sin \\ \sin & \cos\end{matrix}\right]\]
anonymous
  • anonymous
i just do the first term:
rishabh.mission
  • rishabh.mission
that's correct i already done this
anonymous
  • anonymous
a11= cos + tanalpha/2* sin= 1
anonymous
  • anonymous
you can expand the term like what phi says and you get 1 exactly what the left side matrix is
phi
  • phi
for the top right of (2 - sin a + cos a tan(a/2) factor out -tan(a/2) (because that will be the answer) -tan(a/2) ( sin(a)/ tan(a/2) - cos(a)) replace 1/tan(a/2) inside the parens by multiplying by (1+cos(a))/sin a we get -tan(a/2) ( sin(a)*(1+cosa)/sina - cos(a))= -tan(a/2) ( 1 +cosa -cos a)= -tan(a/2) do the same thing for the other entries.
anonymous
  • anonymous
because \[\cos + \sin *\tan \frac{ \alpha }{ 2 }= \cos + \sin\frac{ \sin }{ 1+\cos }= \frac{ \cos +\cos^2 +\sin^2 }{ 1+\cos}\] =1
anonymous
  • anonymous
bingo. phi!! I am with you now. @rishabh.mission is it right?
rishabh.mission
  • rishabh.mission
Just last Query ... how both that matrix same ?
1 Attachment
anonymous
  • anonymous
why do you make that question, why do you separate (I-A) and [cos sin....] matrix to calculate just the latter like that.?you may mess them up.
anonymous
  • anonymous
they are a block and calculate together.
phi
  • phi
I will do the top left entry 1- tan^2(a/2) is 1- sin^2(a/2)/cos^2(a/2) putting over a common denominator we get (cos^2(a/2) - sin^2(a/2)/cos^2(a/2) the numerator is cos(a) (one of version of the identity you need to know or look up) the denominator of the tp left entry 1+ tan^2(a/2) is 1+ sin^2(a/2)/cos^2(a/2) = (cos^2(a/2) + sin^2(a/2)/cos^2(a/2) = 1/cos^(a/2) the cos^2(a/2) cancels and we are left with cos a
phi
  • phi
Can you follow that?
rishabh.mission
  • rishabh.mission
ok i done
rishabh.mission
  • rishabh.mission
@phi i can also use this cos(2a)=1-tan^2a/1+tan^2a
phi
  • phi
that identity is true, but I have not tried to use it to simplify you problem. I tried to explain how I would do you problem.
rishabh.mission
  • rishabh.mission
ok you can solve this using this identity
phi
  • phi
It does not look like a fruitful way to proceed, as you have sin and cos
rishabh.mission
  • rishabh.mission
No we can do using that Eq too
rishabh.mission
  • rishabh.mission
i'll show it to you in morning :)

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