## Aadarsh 3 years ago Trignometry Question

1. Sarkar

Solve: [1/(sin 45 sin 46)] + [1/(sin 47 sin 48)] + ........................+ [1(sin 133 sin 134)]

3. Sarkar

good one...

4. Mani_Jha

The general term of the series is: $t(n)=1/\sin x \sin (x+1)$ Now expand sin(x+1) using sin(A+B). Simplify, and you'll get a really simple general term/

FNSTSE Question

6. Sarkar

i think mani's approach is good!!!!

7. Sarkar

|dw:1334419200461:dw|

So, we get |dw:1334419282831:dw| ?

9. Mani_Jha

Use (sinx)^2=1-cos2x/2 sinxcosx=sin2x/2 Where did sin1 and cos1 go, sarkar?

sin 90 =1, isn't it?

11. Mani_Jha

Actually, sin1=1.17. Yes, aadarsh. You sure you got that general term, Aadarsh?

12. Sarkar

sin1 almost equals 1

Can u say the detailed steps @Mani_Jha bhai and @Sarkar bhai? using all the steps?

14. Mani_Jha

Uh oh, I am not getting a simplified general term using that. Let me think on this.

Sure. Try to post it as soon as possible.

16. Sarkar

note=everyone who have answered this Q is of level 30 and above....lol

I think most of them have left this question. i asked to my teacher, she couldn't answer it.

18. Recursing

http://www.wolframalpha.com/input/?i=sum_%28k%3D45%29%5E133+1%2F%28sin%28k%29*sin%28k%2B1%29%29 something like 1.355867228946

Great!!!!!! Thanks @Recursing . But I want to know the procedure.

20. Recursing

wolframalpha can't find a way to simplify it, if wolframalpha can't I'm sure that it's waaay beyond my level of math knowledge, and I think it's probably impossible to reduce it.

21. Recursing

Oh, wait, i used radians.

@Mani_Jha , @Taufique , @AravindG , @gurvinder bhai, please try once more.

@salini , @shruti , @rebeccaskell94 , @KatrinaKaif please try.

Actually, its a 3 mark question. I just guessed and wrote answer as "1". I don't know how much marks I may have got. But just waiting for over 6 months to get the rite answer. Hope u all help.

@wasiqss , @lgbasallote , @sheena101 , @Akshay_Budhkar , @Ishaan94 please try

@Sarkar , @payalvsangle , @heena , @apoorvk please try.

27. Mani_Jha

I've tried again like this: $1/\sin133\sin134=1/\sin46\sin47$ $1/\sin132\sin131=1/\sin48\sin49$ Similarly, for every term from the right. Finally, $1/\sin91\sin92=1/\sin89\sin88$ Now we'll add these fractions. $(\sin47\sin48..\sin89)+(\sin45\sin46\sin49..\sin50).......(\sin45\sin46\sin47\sin50..\sin89)$ +$(\sin45\sin48..\sin89)(\in numerator)$ Denominator= $(\sin45\sin46.........\sin89)$ Now we'll add the first and last, then second and second last and so on in the numerator. $((\sin48.\sin89)(\sin45+\sin47)+\sin45\sin46..\sin89(\sin47+49)......)/(\sin45..\sin89)$ Now use sinA+sinB=2sin(A+B)/2cos(A-B)/2 $(\sin48..\sin ^{2}88\sin89+\sin45..\sin ^{2}84\sin89...)/(\sin45\sin46...)$ sin88 gets cancelled from the denominator. $(\sin48..\sin89(\sin88+\sin84.....\sin48))/\sin45..\sin89$

So, final answer = ? I could n't understand clearly.

29. Ishaan94

Aadarsh? Is it [1/(sin 45 sin 46)] + [1/(sin 47 sin 48)] + .... + [1(sin 133 sin 134)], or [1/(sin 45 sin 46)] +[1/(sin46 sin47)]+ [1/(sin 47 sin 48)] +...+ [1(sin 133 sin 134)]

30. Ishaan94

The first one

32. Ishaan94

Oh hmm

33. Ishaan94

$\large \sum_{k=0}^{44} \frac1{\sin(45+2k)\cdot\sin(45+2k+1)}$

How tro solve it? What is the final answer?

35. Ishaan94

$\large \sum_{k=0}^{44} \frac{2}{(\sin2k + \cos 2k)(\sin(2k+1) + \cos(2k+1))}$ No idea lol I will have to think about it.

36. Ishaan94

$\large \sum_{k=0}^{44} \frac{2}{\sin(4k+1) + \cos 1}$

37. Ishaan94

This must be a Multiple Choice Question, what are the options?

Subjective Question, 3 marks

39. Ishaan94

I have a feeling something is missing :/ http://www.wolframalpha.com/input/?i= \sum_{k%3D0}^{44}+\frac{2}{\sin%284k%2B1%29+%2B+\cos+1}

40. Ishaan94

@siddhantsharan

41. Taufique

Greta answer @Taufique bhai. Thanks a lot.

43. Taufique

Welcome bhaijan............

45. Taufique

hehe Bhaijan...