- anonymous

Trignometry Question

- jamiebookeater

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- anonymous

AAD-wait for me...

- anonymous

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

- anonymous

good one...

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## More answers

- 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/

- anonymous

FNSTSE Question

- anonymous

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

- anonymous

|dw:1334419200461:dw|

- anonymous

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

- Mani_Jha

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

- anonymous

sin 90 =1, isn't it?

- Mani_Jha

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

- anonymous

sin1 almost equals 1

- Mani_Jha

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

- anonymous

Sure. Try to post it as soon as possible.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

Oh, wait, i used radians.

- anonymous

Well, it doesn't get much better
http://www.wolframalpha.com/input/?i=sum_%28k%3D45%29%5E133+1%2F%28sin%28k*pi%2F180%29*sin%28%28k%2B1%29*pi%2F180%29%29
The partial sums graph looks close to a line, maybe?
http://www.wolframalpha.com/input/?i=sum_%28k%3D45%29%5E133+1%2F%28sin%28k%C2%B0%29*sin%28%28k%2B1%29%C2%B0%29%29

- anonymous

- anonymous

- anonymous

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.

- anonymous

- anonymous

- 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.
\[(\sin48..\sin89(\sin88+\sin84.....\sin48))/\sin45..\sin89\]

- anonymous

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

- anonymous

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)]

- anonymous

- anonymous

The first one

- anonymous

Oh hmm

- anonymous

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

- anonymous

How tro solve it? What is the final answer?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Subjective Question, 3 marks

- anonymous

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}

- anonymous

- anonymous

Greta answer @Taufique bhai. Thanks a lot.

- anonymous

@Aadarsh Welcome Aadarsh ji...

- anonymous

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

- anonymous

hehe Bhaijan...

- anonymous

heheheheheheh

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