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

1. rishabh.mission

See the attachment

2. rishabh.mission

please just solve this Question

3. phi

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

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

5. phi

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

6. rishabh.mission

yes i did

7. rishabh.mission

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

8. phi

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

9. phi

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

10. rishabh.mission

:( not understand

11. phi

or are you asking how they got to that point ?

12. rishabh.mission

yes

13. rishabh.mission

:)

14. rishabh.mission

that's Eq^n 1st

15. phi

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

the matrix above the circled stuff....

17. rishabh.mission

hmm

18. phi

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

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

20. rishabh.mission

sorry ** its $\cos \alpha$

21. Hoa

can I say something?

22. rishabh.mission

yes please .. @hoa

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

24. Hoa

i just do the first term:

25. rishabh.mission

that's correct i already done this

26. Hoa

a11= cos + tanalpha/2* sin= 1

27. Hoa

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

28. phi

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

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

30. Hoa

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

31. rishabh.mission

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

32. Hoa

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

they are a block and calculate together.

34. phi

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

Can you follow that?

36. rishabh.mission

ok i done

37. rishabh.mission

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

38. phi

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

ok you can solve this using this identity

40. phi

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

41. rishabh.mission

No we can do using that Eq too

42. rishabh.mission

i'll show it to you in morning :)