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rishabh.missionBest ResponseYou've already chosen the best response.0
See the attachment
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
please just solve this Question
 one year ago

phiBest ResponseYou've already chosen the best response.0
one way is use the trig identities \[ \tan \left(\frac{x}{2} \right)= \frac{1\cos\ x}{\sin\ x }=\frac{\sin\ x }{1+\cos\ x}\]
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
plz buddy solve this i don't have any idea how to solve..
 one year ago

phiBest ResponseYou've already chosen the best response.0
you could start by writing down IA can you do that ?
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
I have problem with that equation in the circle i shown.
 one year ago

phiBest ResponseYou've already chosen the best response.0
call 2 sin^2(a/2) = x then it looks like 1x+x
 one year ago

phiBest ResponseYou've already chosen the best response.0
**look at the top left expression. Doesn't that simplify?
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
:( not understand
 one year ago

phiBest ResponseYou've already chosen the best response.0
or are you asking how they got to that point ?
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
that's Eq^n 1st
 one year ago

phiBest ResponseYou've already chosen the best response.0
you just posted the left side. on the right side (the messy one) they did IA times the other matrix I take it you understand how they got to the matrix labeled (2) ?
 one year ago

phiBest ResponseYou've already chosen the best response.0
the matrix above the circled stuff....
 one year ago

phiBest ResponseYou've already chosen the best response.0
We could go through that mess, but I would replace tan(a/2) with (1cos)/sin in the top left entry: cos a + sin a tan(a/2) becomes cos a + sin a * (1cos 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
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
can we write \[\cos \alpha/2 = (1\tan ^{2} \alpha/2) / (1+\tan^{2} \alpha/2)\]
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
sorry ** its \[\cos \alpha \]
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
yes please .. @hoa
 one year ago

HoaBest ResponseYou've already chosen the best response.0
left side, i got I+A = \[\left[\begin{matrix}1 & \tan \frac{ \alpha }{ 2} \\ \tan \frac{ \alpha }{ 2 } & 1\end{matrix}\right]\] = right side = (IA) [ ] = \[\left[\begin{matrix}1 & \tan \frac{ \alpha }{ 2}\\ \tan \frac{ \alpha }{ 2 } & 1\end{matrix}\right]\left[\begin{matrix}\cos & \sin \\ \sin & \cos\end{matrix}\right]\]
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
that's correct i already done this
 one year ago

HoaBest ResponseYou've already chosen the best response.0
a11= cos + tanalpha/2* sin= 1
 one year ago

HoaBest ResponseYou've already chosen the best response.0
you can expand the term like what phi says and you get 1 exactly what the left side matrix is
 one year ago

phiBest ResponseYou've already chosen the best response.0
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.
 one year ago

HoaBest ResponseYou've already chosen the best response.0
because \[\cos + \sin *\tan \frac{ \alpha }{ 2 }= \cos + \sin\frac{ \sin }{ 1+\cos }= \frac{ \cos +\cos^2 +\sin^2 }{ 1+\cos}\] =1
 one year ago

HoaBest ResponseYou've already chosen the best response.0
bingo. phi!! I am with you now. @rishabh.mission is it right?
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
Just last Query ... how both that matrix same ?
 one year ago

HoaBest ResponseYou've already chosen the best response.0
why do you make that question, why do you separate (IA) and [cos sin....] matrix to calculate just the latter like that.?you may mess them up.
 one year ago

HoaBest ResponseYou've already chosen the best response.0
they are a block and calculate together.
 one year ago

phiBest ResponseYou've already chosen the best response.0
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
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
@phi i can also use this cos(2a)=1tan^2a/1+tan^2a
 one year ago

phiBest ResponseYou've already chosen the best response.0
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.
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
ok you can solve this using this identity
 one year ago

phiBest ResponseYou've already chosen the best response.0
It does not look like a fruitful way to proceed, as you have sin and cos
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
No we can do using that Eq too
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
i'll show it to you in morning :)
 one year ago
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