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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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