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rishabh.mission
 2 years ago
matrix Question ....plz help
rishabh.mission
 2 years ago
matrix Question ....plz help

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rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0See the attachment

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0please just solve this Question

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

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0plz buddy solve this i don't have any idea how to solve..

phi
 2 years ago
Best ResponseYou've already chosen the best response.0you could start by writing down IA can you do that ?

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0I have problem with that equation in the circle i shown.

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.0**look at the top left expression. Doesn't that simplify?

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0:( not understand

phi
 2 years ago
Best ResponseYou've already chosen the best response.0or are you asking how they got to that point ?

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0that's Eq^n 1st

phi
 2 years ago
Best ResponseYou've already chosen the best response.0you 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) ?

phi
 2 years ago
Best ResponseYou've already chosen the best response.0the matrix above the circled stuff....

phi
 2 years ago
Best ResponseYou've already chosen the best response.0We 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

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

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0sorry ** its \[\cos \alpha \]

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0yes please .. @hoa

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

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0that's correct i already done this

Hoa
 2 years ago
Best ResponseYou've already chosen the best response.0a11= cos + tanalpha/2* sin= 1

Hoa
 2 years ago
Best ResponseYou've already chosen the best response.0you can expand the term like what phi says and you get 1 exactly what the left side matrix is

phi
 2 years ago
Best ResponseYou've already chosen the best response.0for 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.

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

Hoa
 2 years ago
Best ResponseYou've already chosen the best response.0bingo. phi!! I am with you now. @rishabh.mission is it right?

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0Just last Query ... how both that matrix same ?

Hoa
 2 years ago
Best ResponseYou've already chosen the best response.0why 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.

Hoa
 2 years ago
Best ResponseYou've already chosen the best response.0they are a block and calculate together.

phi
 2 years ago
Best ResponseYou've already chosen the best response.0I 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

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.0that 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
 2 years ago
Best ResponseYou've already chosen the best response.0ok you can solve this using this identity

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

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0No we can do using that Eq too

rishabh.mission
 2 years ago
Best ResponseYou've already chosen the best response.0i'll show it to you in morning :)
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