## rishabh.mission Group Title matrix Question ....plz help one year ago one year ago

1. rishabh.mission Group Title

See the attachment

2. rishabh.mission Group Title

3. phi Group Title

one way is use the trig identities $\tan \left(\frac{x}{2} \right)= \frac{1-\cos\ x}{\sin\ x }=\frac{\sin\ x }{1+\cos\ x}$

4. rishabh.mission Group Title

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

5. phi Group Title

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

6. rishabh.mission Group Title

yes i did

7. rishabh.mission Group Title

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

8. phi Group Title

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

9. phi Group Title

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

10. rishabh.mission Group Title

:( not understand

11. phi Group Title

or are you asking how they got to that point ?

12. rishabh.mission Group Title

yes

13. rishabh.mission Group Title

:)

14. rishabh.mission Group Title

that's Eq^n 1st

15. phi Group Title

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) ?

16. phi Group Title

the matrix above the circled stuff....

17. rishabh.mission Group Title

hmm

18. phi Group Title

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

19. rishabh.mission Group Title

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

20. rishabh.mission Group Title

sorry ** its $\cos \alpha$

21. Hoa Group Title

can I say something?

22. rishabh.mission Group Title

23. Hoa Group Title

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) [ ] = $\left[\begin{matrix}1 & \tan \frac{ \alpha }{ 2}\\ -\tan \frac{ \alpha }{ 2 } & 1\end{matrix}\right]\left[\begin{matrix}\cos & -\sin \\ \sin & \cos\end{matrix}\right]$

24. Hoa Group Title

i just do the first term:

25. rishabh.mission Group Title

that's correct i already done this

26. Hoa Group Title

a11= cos + tanalpha/2* sin= 1

27. Hoa Group Title

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

28. phi Group Title

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.

29. Hoa Group Title

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

30. Hoa Group Title

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

31. rishabh.mission Group Title

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

32. Hoa Group Title

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.

33. Hoa Group Title

they are a block and calculate together.

34. phi Group Title

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

35. phi Group Title

36. rishabh.mission Group Title

ok i done

37. rishabh.mission Group Title

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

38. phi Group Title

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.

39. rishabh.mission Group Title

ok you can solve this using this identity

40. phi Group Title

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

41. rishabh.mission Group Title

No we can do using that Eq too

42. rishabh.mission Group Title

i'll show it to you in morning :)