matrix Question ....plz help

- rishabh.mission

matrix Question ....plz help

- schrodinger

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- rishabh.mission

See the attachment

##### 1 Attachment

- rishabh.mission

please just solve this Question

- 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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## More answers

- rishabh.mission

plz buddy solve this i don't have any idea how to solve..

- phi

you could start by writing down
I-A
can you do that ?

- rishabh.mission

yes i did

- rishabh.mission

I have problem with that equation in the circle i shown.

##### 1 Attachment

- phi

call 2 sin^2(a/2) = x
then it looks like 1-x+x

- phi

**look at the top left expression. Doesn't that simplify?

- rishabh.mission

:( not understand

- phi

or are you asking how they got to that point ?

- rishabh.mission

yes

- rishabh.mission

:)

- rishabh.mission

that's Eq^n 1st

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- 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

the matrix above the circled stuff....

- rishabh.mission

hmm

- 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

can we write \[\cos \alpha/2 = (1-\tan ^{2} \alpha/2) / (1+\tan^{2} \alpha/2)\]

- rishabh.mission

sorry ** its \[\cos \alpha \]

- anonymous

can I say something?

- rishabh.mission

yes please .. @hoa

- 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

i just do the first term:

- rishabh.mission

that's correct i already done this

- anonymous

a11= cos + tanalpha/2* sin= 1

- anonymous

you can expand the term like what phi says and you get 1 exactly what the left side matrix is

- 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

because \[\cos + \sin *\tan \frac{ \alpha }{ 2 }= \cos + \sin\frac{ \sin }{ 1+\cos }= \frac{ \cos +\cos^2 +\sin^2 }{ 1+\cos}\] =1

- anonymous

bingo. phi!! I am with you now.
@rishabh.mission is it right?

- rishabh.mission

Just last Query ... how both that matrix same ?

##### 1 Attachment

- 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

they are a block and calculate together.

- 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

Can you follow that?

- rishabh.mission

ok i done

- rishabh.mission

@phi i can also use this cos(2a)=1-tan^2a/1+tan^2a

- 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

ok you can solve this using this identity

- phi

It does not look like a fruitful way to proceed, as you have sin and cos

- rishabh.mission

No we can do using that Eq too

- rishabh.mission

i'll show it to you in morning :)

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