I bet you can't solve this IIT JEE Advanced type question
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- Опубліковано 21 бер 2024
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Bhaiya I just entered class 11(jee aspirant)
Any suggestions you would like to give me ...
Answer galat hai kyuki sine(pi/n) sin(2pi/n) sin((2pi(n/2-1) =√(n/2^n-1). Derivation mai n/2-1 two bahar ayega naki n/2 aur lamda =3
Hogya bhaiya kyoki maune complex me ye padha tha
Product wala chiz toh maine waha se formula wagera derive krliya
Exactly@@abhinavanand9032
Can you make advanced level playlist for complex numbers as it is hard to find any good video for complex ??? Plz
lambda = 3 will be the answer ( can check from desmos using Pi function)
i did it from trigonometry -
sin1 sin2 ..... sin90
= sin1 cos 1 sin2 cos2 ... sin88 cos88 * sin(45) [will be using the sin45 at last]
=1/2^44 (sin2 sin4 sin6....sin88)
similarly this will be equal to -
1/2^66 (sin4 sin8 sin12 ... sin88) [all multiples of 4]
= 1/2^66 (cos2 cos6 cos 10 ... cos86) [just converted sin to cos]
now just form pairs of 3 to use the idenitiy cos(A)cos(60+A)cos(60-A) = 1/4 cos3A
=1/2^66 *cos30 [ cos2 cos62 cos58 * cos6 cos66 cos54 .....]
= 1/2^80*cos30 [cos6 cos18 cos30 cos42 cos54 cos66 cos78]
apply the same identity again -
1/2^84 *cos30 *cos30 [cos18 cos54]
now cos18 * cos 54 = root80 /16 and (cos30)^2 = 3/4 and sin45 = 1/root 2
plug in all these in the final equation
= 3 (root40) /2^90 = lambda root10 /2^89
we get lamda = 3
not a very lengthy procedure if one remembers the formulas and values like sin18 sin36 etc
same approach kari :)
Bhaiiiiii 🫡
Salute
Cos3A ki identity lagate samay atak gya nice solution
Wow very smart😎
Youre doing an excellent job. I just want to say if you put the question in title, if somebody does search for question, he/she can find it. Bcoz the question you put has practically no understandable video in UA-cam, so that will be very very helpful ❤
Bro is obssesed with complex numbers
Hardest (university level) chapter for a reason 🔥
i disagree@@rishuvaiya1305
Bhai spoiler mat dia kar pls😢
Though I don't know the name of the person who is making these videos, but I am sure that he writes his name like the following....COS (first name) + i SIN ( last name ).
Bhaut easy hai bhai @rishuvaiya1305
I wasn't able to solve it but was sure that it will involve some application of nth roots of unity... Really enjoyed this one👍
Bhaiya I have discovered you fairly recently and I love the questions you present. Just one request please attach some questions at the end so that we may attempt it ourselves . Thank you
Bhaiya aap physics aur chemistry ke liye bhi aise hi badhiya questions laya karo please..
Heyy itz crazy nd thnk u bhai
@JEE SIMPLIFIED SUBJECT Can you please make a short video explaining this roots property at 1.53 It would be really helpful!
Use this standard result and get answer in 1min
Sin(π/n)sin(2π/n)sin(3π/n).....sin (n-1)π/n= n/{2 ^(n-1)}
Actually this result is obtained by using exactly same approach which is using in this video.
Toook me 13 min 2 ki power 89 se click kiya ki ismey pehle sin ko club karnege sin2x use karke phir cos me convert karke series lagayi aur pehle se hi upon me 2 ki power jo extra multiply ki haj wo bacchi rhi uske baadey lapet diya aur agr iss method se kre toh itna time lg hi jana kyunki lst me cos ki values nikal ke put kiya mainey yaad nhi thi sorry abb 54 18 ye sbb bhula jaata😊😅
Bhaiya same proof I did geometrically
As nth roots of unit all of them lie on a circle and area space tetha = 2pi/n apart
now take the first root and 1 in a circle draw a line connecting them (chord of the circle)
now using basic circle properties the length of a chord of a circle subtending angle tetha would
be 2Rsintetha/2 and as tetha is 2pi/n here 2R sin(pi/n) will be the length of the chord
R is = to 1 therefore length of chord is is 2 sin(pi/n) and chord complex number wise can be
represented as (e^i2pi/n - 1) so mod of it will be equal to 2sinpi/n
similarly (e^i4pi/n - 1) mod of that is 2 sin2pi/n
so onn til (e^i (n-1)pi/n - 1) mod of that will be 2 sin (n-1)pi/n
so multiply all terms both sides u get
2^n-1 (sinpi/n * sin 2pi/n .....sin(n-1)pi/n) = |1-e^ipi/n| * |1-e^i2pi/n|*.....|1 - e^i(n-1)pi/n|
so the rhs of prec equation is same as keeping x=1 in equation (x^n - 1)/(x -1)
using this result i took n = 180 used symmetry and square rooted and got the ans
Credits to my sir for sharing this solution 🙏
Notes questions and challenge is to do with cosine series
Nice question
Bro pls tell what books should i do for maths i think arihant is good but then think it is legnthy and 1year is suff or not for that and if i think the black,yellow,pink books by vikas gupta then they seem to be revision oriented more rather than for practicing ch for first time cengage has the same problems has arihant and other books are just not complete pls help
Ye saare AIR 1-100 isi comment section mai hi milte hn. Sabh koi hoshiyaari maar rhe hn. Real mai inse mains ke question nhi hote honge
bhai tu samajhdaar lag raha h.
take this word of advice.. apne aap pe focus kar. koi bhi random sawaal solve karne mat bethna kabhi. pyqs 2008-present saare karle
give all of them as mocks if possible and apne marks dekh. jee advanced ki baat kar raha hu. agar 45-50% ya usse zyada hai. you are good to go. koi kuch bi bol le comment section mein. statistically proven hai ye fact less than 0.2% of students can score more than 45% marks. pichle 4-5 saal ka data utha ke dekh lena.
approx yahi reality hoga.
@@godson200 haan 👍👍
Tum 12 class mai ho?
@@shivam2694 mein 4th year m hu, iit
@@godson200 acha. Nice 👍
@@ptwsingh iit bombay shayad phod bhi de... lekin textile 🫡🫡
Sone to diya karo, last moment pe video daal dete ho
ab dekhni padhegi
usne teri gardan pakad ke rkhi hai 🤡
Solved in 10 mins, Aese similar Summation CMI ke subjective part me dekha tha shayad thode saal pehle usliye help hua
Alternative method using trigonometry using SinA sin(60-A)sin(60+A)=
(1/4)sin3A
Put A= 1,2,3,......30°
And multiply all these 30 equations,
Now put A=3,6,9,.... 30°
And multiply all these 10 equations and so on
You can do by comolex also
Bhai apka samjhaya hua que class me jhadta hu ab teacher topper samjhta hai, kya karu?
❤❤
*"Bhai thoda vo 2nd term aur last term ko multiply karneka logic batao na, nahi toh maths bhi bio ki tarah ban jayega😢"*
ans would be 3 also u wouldnt need to multiply 1st and last term put z as 1 and it will form 1-cos2mpie+isin2mpie and here take common 2sinmpie and take mod it will only left with 2 sinmpie now genralise to our identity it will form (2^89)*question = gp (with z=1) hence it will reduce to 90/2^89 hence ans will be 3
aur ha maine ye phele kiya hua tha
Bhot specific result yaad tha lol
Ke un sab ka multiplication gives 6root10/(4^45) to udhar se hogaya
Par solution bhi bhot achha tha
I saw this problem on allen digital as an illustration
Solved by trigonometry and got lambda = root(1/5)
Lamda 3 aarA hai sir uski value 180by 2 ki power 179 aati hai
I GOT THIS q bhaiya our sir already made us solve many multiplicative series using complex numbers so i hadnt had to think. outside the box i knew complex number se kuch that and i fiddled around and got it i took 3 min
(if my sir didnt give a similar q before i wld have taken a loooot more time)
I loved the line *Bade se Bade gyani ki band bajane wala* 🔥
Question be like : 😎🍭
Directly using the product of sine formula we get 3
The square of the given product can be calculated using the formula n/2^(n-1)
hi do you have aany app where i can cotanct you i wanna know how to use that formula
3/root(2) done in around 6 minutes done these type of questions in class of complex
3 aana chaiye
2024-25 11TH BATCH
PLEASE somebody tell do we need separate copies for coaching and school (regular btw) ??
This variety of question was in my class notes😂😂
How could we derive cos series
Good Question with Nice Solution
🙏
bhaiya yeh q hmare coaching mai choc problem ke roop mai diya tha….jab bhi nhi hua and abhi bhi nhi
Sine AP formula (derived from de moivres)
Continued product of sin(rπ/n), as r ranges from 1 to n-1 is n/2^(n-1). Place n=180 and simplify, great result used in this question
ye formula exam ke din yaad rakh paoge??
@@ClarkKent-bz9tf obviously Maine bohot practice kiya hai is formula pe kyu nahi rkh paunga agar physics ke itne bade bade yaad rkha skta hu toh yeh tho bacchha hai uske aage
@@ClarkKent-bz9tf bro i learned it by 2n/2^n its more elegant to learn that way, and I have it memorized now because it has a very beautiful derivation
@@Ayush-mg6xw ok bro but mai personally apne pr itna trust nhu krta aur honestly maine ye formula kabhi suna hi nhi tha and in this question i used the same process jo formule ko derive krne mai lagta , uske baad maine genral n/2^(n-1) wala format derive kia aur waise bhi kis bhi achhi book mai trigno ki series pr aise 100s of question hai jo saare different hai sab ka formula to yaad nhi rakh paayega for example agar mai tujhse puchu ki tan(A/2) ka formula in terms of semiperimeter kya hoga to tujhe yaad hai? isiliye mai personally formule ke peeche ke ideas ko yaad rakhta hu
@@ClarkKent-bz9tf yaad hai na bhai √(s-b) (s-c) /s(s-a))
Bhai tujhe kya lagta hai top rankars adhe se jyada question rate hue hote hai aur results toh aise pata hote hai joh tune sochi bhi nahi hogi jaise what is a solid angle of a rectanglular surface isko derive krta zindagi kharab ho jayegi kyunki apne syllabus mein bhi shyad hai but formula yaad hoga tabhi question kar payega baaki i agree logic ana chahiye voh toh mai bhi keh rha hu lekin formula are very very very important
bhaiya lamda =3 hoga ans ....... n/2^(n-1) = | sin(pi/n) sin 2pi/n sin 3pi/n .... sin(n-1)pi/n | result yaad tha ,,, 11th mai derive karwaya tha then yaad karne ko bola tha. ... root 180/2^179 = root90 / 2^89 = 3rt10/2^89
Guessed the complex number but after few hints got it 😊
As I have know the formula already
Bhaiya aap mera email kiya hua question kab dekhoge 😢😢
After you ll give JEE
Same mera bhi nhi dekha
Wapas se bhejde bhai , bahut questions aate hoge
Aisa hi same question sir ne 11th mae class mae kraya tha..isliye ye toh hogya 5 min lagi🤘
Konse sir konsi coaching...?
It's simple multiplication both side by Sin0
nice!
Anya should study in lower classes not JEE 😀
Bhaiya answer 3 aayega
Direct result hota h complex number se
sin(x)sin(2x)sin(3x)....sin(n-1)x= n/(2^n) where nx=pi
To isse use kare to lambda 3 aata h
Bhai abhi abhi padha thaa lekin ye question dekh ke nahi soch paya :(
@@Kirtiraj_Deshmukhbro how tf did you not see it was the question made for this formula
Even though it was a bit rigorous,
I solved it by trigonometry because I have done a question which helped me here.
First of all we have 89 terms (sin 90 = 1) so I took sin45 away and made pairs of (theta) and (90-theta). I multiplied and divied by 2^44 and applied the formula 2sinAsinB to all pairs.
2sinAsin(90-A) would just give cos(2A-90)
So now the series converted into
(1/2^44) Multiplied by cos2•cos4•cos6 ... Upto cos88
Now If you've solved lots of trigonometry questions you'd see that here cos(theta) , cos(60+theta), cos(60-theta) triplets would form.After using sin(theta)•sin(60+theta)•sin(60-theta) = 1/4 sin(3theta) and simplifying all triplets you will end up with (1/2^44) × (1/2^40) × cos18 cos54 cos30 sin45 which upon solving would give the right value of lamda.
It might seem really long but It took me around 4-5 minutes to solve it since I knew the exact way to solve.
Bhaiya's method of solving was also good ❤
How is 2sinAsin(90-A) = cos(A-B) ??
@@MayurAvad 2sinAsinB = cos(a-b) - cos(a+b)
In this case a+b = π/2 so cos(a+b) = 0 so only cos(a-b) remains. I forgot to put the value of a and b in the comment. I have fixed it now
Bro genius..I was too scared to solve it by rigorous trigno..you opened my eyes
@@yashkumar2327 just have confidence in yourself bro that's what I did
@@rawatutkarsh yeah now I get it..I also thought of solving by this method by first glance but was too lazy to solve at midnight 😂 I liked this soln more tho
Same ques in MT sir notes
we can also solve it by product of cosine series
No
Isme ham sin x ko imz(e^ix ) likhe aur sab terms ko multiply karke powers add karke answer nhi nikal sakte ? Usse answer alag aa raha hai
im(z1)*im(z2) zaruri nhi hai ki im(z1*z2) ho, toh nhi waha se galat answer aayega
SinA=(e^iA-e^-iA)/2i likho dono side modulus lele lo aur phir simplify karo toh aisi series ka ek general formila a jata hai
Yes...and that might be the shortest way.
@@timirbiswas6785 the formula for such a Series is n/(2^(n-1))
Any idea about cos series
@@TGRRohit same method
solved in 20 seconds i am not joking... i am class 11 student i derived a general formula for product of sines when learning complex numbers as i was very fascinated by it and this was just a direct application of it :) i have derived many weird results using complex numbers for solutions of trigonometry. also i got 3 as the answer
Same bro
isse koi fayda nhi hoga...
@@eobardthawne19 bro agar derivation coaching mai indepth karra diya ho toh fir usko baar baar thodi karenge, sirf application yaad rakhenge. I know derivations are important, thoda question change kar diya toh tab derivations hi kaam ke aate, lekin ye wala direct tha
Mind sharing a piece of your effort regarding some results i will show some mine as well...
@@Maverick2110hi bro I am curious how did you derive those you have any videos reding this ?
bro aapka mistake samj me aagaya jab aab 2^n/2 common lerahe the (second time from 2-2cosrπ/n)ek 2 nahi aayega sirf 2^(n/2 -1) common aaya tho wo pura chiz √n/2^(n-1) bhan jayegi so abb n=90 daalne se sahi answe lamda 3 aajayega
Bhai sone do na yar 😅
Distract mat karo na ab dekhna padega
fittz3e wale aits me same aaya tha
Pehli Nazar me to complex laga ab age solve karke dekhta hu
Bana to liya tha par notes ka help leke !!!
1 subscribe = 97.69% in boards
Ignore = 32.69% in boards, risk liya to shrap dunga💀
There IS definitely some error bhaiya.
x^n-1 k n roots honge na
Unme se x-1 ko hum LHS mein le gaye
To n-1 hi bache iss side
So 2^n common nahi aayega
Also n even odd kya rahega hum khud figure out karenge because hume pata hai 1 se 90° tak nikalna hai
bhai bhaiya ne n = pi dalne ke bad radian se degree mai kaise covert kiya?
Aiats mein aaya tha aaj similar question
Bhaiya, mai abhi 11th mai hu, coaching mai abhi ek mahine pehle hi complex numbers karaye the toh problem dekhte hi sin wali identity yaad aa gyi jo sir ne bade depth Mai class mai padhai thi. Bas calculation aur setting karna tha, toh ye wala questions honestly 1 min ho gya. Maine baki vids mai cmnt nhi Kiya tha kyuki unme more than 10 mins lage the lekin ye wala ho gya easily. Answer is λ=3
Voh complex number wale na where we used Euler identity and nth root of unity to derive the general solution
@@WilliamJamesMoriarty-hf6re yes
Bhai I solved it by.....converting all sine after 45 to cos.....and applying 2sinacosb wala formula....to get an ap in sine...whose formula we already know
Exactly ye qn class illustrations mein v hota hai
Bhaiya pehli baar aasan question upload krdiye aap to
Ye to direct result h nth roots ka jo worth h yaad rkhna
Product sinkpi/n from 1 to n-1 =n/2^n-1
assume the product in ques to be p
p is also equal to sin91sin92sin93……sin179
P^2=product sinkpi/180 from 1 to 179
Now take sqrt both side and solved
Me JEE nahi kar raha hu mujhe maths bohot pasand hai isiliye ye dekhta hu.
Bhaiya agar sin1sin89=sin2/2 aise pairs banayenge to ho rha h
Ha correct
Done this before iska square is 180/2¹⁷⁹( nth root of unity kaa result)
From fiitjee notes 💀
Ye question dekhte mai samajh gayi thi ki aap kyu kah rahe ho ki ye trigonometry ka nahi hai.
Actually kuch hafte pehle jab hum complex number padh rahe the to sir ne hume general formula nikalne ko kaha tha for ccos and sin jab cos ke angle AP me ho aur unka product ho raha ho .
Uss time mera sir bhi ghooma tha😅
Cos me covert karke sin1 se multiply divide kar sakte?
No
me thinking that ~ sin(0) bhi hona chaiye tha starting mai 😁
Bhaiya I have solved this
People in commnt section👽
People in real exam🥲
Jee simplified wife reveal - complex numbers
Comment section jhak ke dekhta hu shayad koi hint mile😅
sir mistake hui hai
ans 3 aana chahiye
ek baar check karo
Maja aa gaya
Kaash Sin0 bhi aage multiply kar diya hota ❤
now i will get nightmares
You can solve from trigonometry as well just take this series to be x and another series y analagous to this that is cos1°×cos2°........×cos(89°) then multiply both of them by some trig transformationa you get the same series y they cancel out and you get the series x in terms of the factors you used to convert xy as a linear function of y 1/2^something etc. generally people can solve this is a very standard problem everybody who is not an idiot knows this so nothing special bring something which is actually challenging you keep bringing standard problems
Hey I have solved this question using complex numbers but seeing your method it looks easy but I don't understand your explanation and you explain it properly
@@garoutheherohunter8613 just multiply and divide by the cosine of the corresponding angles and the same number of 2's as well to get the series of the form (sin(2theta1)×sin(2theta2)×sin(2theta3).....sin(2theta'n'))/((2^n)×(cos(theta1)×cos(theta2)......cos(theta'n')) ,now with the set of number you have been given in the question you write theta(1) as pi/2-theta(i) for some 'i' you need to find this 'i' yourself others will follow the same pattern and your numerator 'sine' series converts to the cosine series of the denominator and the cosines cancel out .Try this out yourself! You will love it😀 you can also see the unacademy Jaipur pulse video of a maths teacher I can't remember the name of ,he was teaching the same trigonometric concept wait I will write it in editing
@@garoutheherohunter8613 it's pankaj baluja sir Jaipur pulse by unacademy trigonometric product of angles in arithmetic progression
@@user-sr1he9fp1s thanks I saw his vid this method was very easy whereas I was using Euler formula for solving this question that sin x = e^ix - e^-ix but this method is very easy but Euler method helps us to get a general solution for these type of problem that why I recommend you to check it out
sirdard karwa di bhaiiiiiii merese na hota ye sb
Kvpy pyq2021
Aur competition badhao please🤜🤛⬛
Can anyone please suggest books for *maths* for IIT JEE... I just got into 11th std and will be doing online classes. Can any senior pls help me!! Can y'all pls suggest some maths reference books for it 🙏🙏
cengage. bas kaafi hai. ye khatam hoga uske baad yaha coment section m aana aur book ke liye.
aur ncert solve karo pehle toh fir cengage khatam karo. examples illustrations exercises.
koi black book, pink book, kuch zarurat nahi h. bas cengage start to end khatam karo. aur ho sake toh you can do pyqs of jee mains side by side with each chapter.
I REPEAT MYSELF DONT GO FOR ANY OTHER BOOK. I AM ASSUMING YOU DONT HAVE ANY COACHING MODULE HENCE SUGGESTING THE NEXT BEST THING. STICK TO ONE MATERIAL.
@@godson200 okayyyyy! Got it, I will surely do the same. This is actually what I needed to know n confirm!! Tysm
Bhaiya Lambda = 3 hona chahiye
Me who just entered class +1 and saw this
Main to gaya bahiya
Aapne to sonee ko bola tha 11:30 tak 😅
3 hi aayega answer
Bhaiyaa koi aur approach nahi hai kya
Trigonometry vagera se
😢😢😢
Aaj shah rukh khan nahi dikha :(
Bro trying to solve jee mains question by jee advanced solutions😅
Bhai kal hi advance ke test m aaya tha 😂😂
Maine tukka lgya aur shi ho gya 😅😸
ye wala smjh nhi aya
Kal subha karunga am tired rn
bhaiya ho gya. aur ek method se. i mean sin 89ko cos 1 likhke ulta karke kia
hogya, bet haar gaye ab mere paise kha hai (aur ans 3 hai naki 3/root2)
Iska mereko formula pdhaya thaa...
Direct method❤
Mereko batana BHai Vo mEthod Plz
Aur agar kar diya to kya doge? Is month ka revenue🌝🌚
Saare Air 1 isi comment section me milte hai 😂
Bhai 3 aa raha hai na lambda?
same mera bhi 3 aya