Here's the question you clicked on:
rishabh.mission
matrix Question ....plz help
See the attachment
please just solve this Question
one way is use the trig identities \[ \tan \left(\frac{x}{2} \right)= \frac{1-\cos\ x}{\sin\ x }=\frac{\sin\ x }{1+\cos\ x}\]
plz buddy solve this i don't have any idea how to solve..
you could start by writing down I-A can you do that ?
I have problem with that equation in the circle i shown.
call 2 sin^2(a/2) = x then it looks like 1-x+x
**look at the top left expression. Doesn't that simplify?
:( not understand
or are you asking how they got to that point ?
that's Eq^n 1st
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) ?
the matrix above the circled stuff....
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
can we write \[\cos \alpha/2 = (1-\tan ^{2} \alpha/2) / (1+\tan^{2} \alpha/2)\]
sorry ** its \[\cos \alpha \]
yes please .. @hoa
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) [ ] = \[\left[\begin{matrix}1 & \tan \frac{ \alpha }{ 2}\\ -\tan \frac{ \alpha }{ 2 } & 1\end{matrix}\right]\left[\begin{matrix}\cos & -\sin \\ \sin & \cos\end{matrix}\right]\]
that's correct i already done this
a11= cos + tanalpha/2* sin= 1
you can expand the term like what phi says and you get 1 exactly what the left side matrix is
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.
because \[\cos + \sin *\tan \frac{ \alpha }{ 2 }= \cos + \sin\frac{ \sin }{ 1+\cos }= \frac{ \cos +\cos^2 +\sin^2 }{ 1+\cos}\] =1
bingo. phi!! I am with you now. @rishabh.mission is it right?
Just last Query ... how both that matrix same ?
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.
they are a block and calculate together.
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 i can also use this cos(2a)=1-tan^2a/1+tan^2a
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.
ok you can solve this using this identity
It does not look like a fruitful way to proceed, as you have sin and cos
No we can do using that Eq too
i'll show it to you in morning :)