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Trignometry Question

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AAD-wait for me...
Solve: [1/(sin 45 sin 46)] + [1/(sin 47 sin 48)] + ........................+ [1(sin 133 sin 134)]
good one...

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Other answers:

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
i think mani's approach is good!!!!
So, we get |dw:1334419282831:dw| ?
Use (sinx)^2=1-cos2x/2 sinxcosx=sin2x/2 Where did sin1 and cos1 go, sarkar?
sin 90 =1, isn't it?
Actually, sin1=1.17. Yes, aadarsh. You sure you got that general term, Aadarsh?
sin1 almost equals 1
Can u say the detailed steps @Mani_Jha bhai and @Sarkar bhai? using all the steps?
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.
note=everyone who have answered this Q is of level 30 and
I think most of them have left this question. i asked to my teacher, she couldn't answer it.*sin%28k%2B1%29%29 something like 1.355867228946
Great!!!!!! Thanks @Recursing . But I want to know the procedure.
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.
Oh, wait, i used radians.
Well, it doesn't get much better*pi%2F180%29*sin%28%28k%2B1%29*pi%2F180%29%29 The partial sums graph looks close to a line, maybe?*sin%28%28k%2B1%29%C2%B0%29%29
@Mani_Jha , @Taufique , @AravindG , @gurvinder bhai, please try once more.
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.
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.
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)]
The first one
Oh hmm
\[\large \sum_{k=0}^{44} \frac1{\sin(45+2k)\cdot\sin(45+2k+1)}\]
How tro solve it? What is the final answer?
\[\large \sum_{k=0}^{44} \frac{2}{(\sin2k + \cos 2k)(\sin(2k+1) + \cos(2k+1))}\] No idea lol I will have to think about it.
\[\large \sum_{k=0}^{44} \frac{2}{\sin(4k+1) + \cos 1}\]
This must be a Multiple Choice Question, what are the options?
Subjective Question, 3 marks
I have a feeling something is missing :/\sum_{k%3D0}^{44}+\frac{2}{\sin%284k%2B1%29+%2B+\cos+1}
@Aadarsh read the attach file
1 Attachment
Greta answer @Taufique bhai. Thanks a lot.
@Aadarsh Welcome Aadarsh ji...
Welcome bhaijan............
hehe Bhaijan...

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