Long but predictable method: sin (kpi/m), k=1..n are roots of equation sin(mx) = 0. So we expand sin(mx) in powers of sin(x) to get a polynomial equation in sin(x) of degree m. For m = 7 we get: sin(7x) = -64sin^7(x) +112 sin^4(x) -56 sin^2(x) + 7sin(x) =0. We can eliminate sin(x)=0 as solution, so the products of the remaining 6 roots is 7/64. This product is square of the required quantity.
Yeah bro same 😂 I once was curious and derived sin5x , 7x 9x 11x and those r some neet results if u want an interesting question check this .. what is the coefficient of last 3rd term of sin101x given that it's expansion is organised on the basis of descending order of powers of sinx ... Sayd question smjh me aagya hoga kya bol rha .... Actually mein general term ka coefficient khoj rha tha nahi mila but kuchh terms me pattern mila
Alternate method is by complex No z⁷ = 1 ( Roots are 1 , x,x².... x⁶) x and x⁶ , x² and x⁴ ,x³ and x⁵ are conjugate Now z⁷-1 =(z-1)(1 -x)(1-x²)...(1-x⁶) Collect the above pairs that are conjugate and simplify z⁷ -1=(z-1)( z² - 2Re(x) +1)..(z² - 2Re(x³) +1 ) Put the values in cos form of Re(x) and put z=2
Sir, i saw your video on the same problem and the next morning when i was searching for the video, i didn't find it anywhere, not even in my history, but then i thought that you might have deleted the video. Thanks for re uploading the problem.
In general , it can be proved using complex numbers that: multiplication of 'N' terms of sines where each acute angle is in A.P. such that the first term is equal to common difference and last term is just less than π , is equal to (N+1))/2^(N). In the question in this video , let the value to be determined be X . Then , X²= sin(π/7)sin(π/7)sin(2π/7)sin(2π/7)sin(3π/7)sin(3π/7) Now using sin(x) = sin(π-x) , we have , X^2 = sin(π/7)sin(2π/7)sin(3π/7)sin(4π/7)sin(5π/7)sin(6π/7) The last equation follows the formula above , hence , X^2 = 7/2^6 = 7/64 . Take square root and you get your answer.
Hello :) I wanted to know how you came to know about this identity, because I am halfway through my 11th std., starting the chapter 'Circles' tomorrow, and I never came across such an identity while studying the two trigonometry chapters in the past. Thanks.
Sir this question could be done easily by using the concept nth root of unity, Where n is 7. (x^7-1)/(x-1)=1+x+x^2+...+x^6=(x-e^i2pie/7)....(x-e^i12pie/7) Club conjugate vise all the 6 factors in rhs. They will become 3 , (x^2 - 2xcos(2pie/7) +1)... Put x=1 Lhs = 7 Rhs = (8)(1-cos(2pie/7)).... And rhs = 64(sin(pie/7).sin(2pie/7).sin(3pie/7))^2 So got it
Well explained sir!! I appreciate your effort 😊👍 But sir ,,i have an another excellent approach,,which is luckily a brilliant and time saver approach for this particular question!! We have a formula ,,,sinπ/n .sin 2π/n .sin3π/n ... .sin (n-1)π/n = n/{2^(n-1)}. In this particular problem we have value for sinπ/7 sin2 π/7 ...sin6π/7 = 7/(2^6).. Now we let us assume sin π/7...sin3π/7 =X .then the next three terms can be transformed in terms of Series of X.as eg:-sin4π/7 =sin 3π/7..now the series become X^2 in LHS...now apply root both side .we get value as √7/8.. Its so simple,,can be done within 4-5 lines😊😎🔥🤘 Lets have a look to my approach! Hope u like it !!. Thankyou! Radhe radhe🙏
@@shauryagupta3644i appreciate your curiosity 😊!! And thankyou for read my approach 😊 . To be honest ,,this formula is not related to intermediate level directly!! And the proof is so lengthy ,you have to use eulers identity ,concept of limit ,and little bit use of complex no. Yeah iam also an iit jee aspirant right now😊 so ,no doubt you can understand the whole proof almost ! But this will worthless for u right now !! The result is important ,,so plz try to keep that result in mind ! Remember,,you cant use this formula in every situation,,the major condition is the coefficient of π in last term is 1 less than the denominator !!and the product should be continuous!! Further modification in this formula led to the result of "Riemann integral ". But still you want to know the proof ,,i have done an effort in searching the video you need to get the proof . Iam providing you link ,,you can chek it out!😊👇 ruclips.net/video/hQ532SjbxNo/видео.htmlsi=auUbzdDu8AipVGw_ This contain the proof u want,,and given by any university professor !!👍. Once again ,i say ,,result is important!! Thankyou to love my approach!!😊👍 Radhe radhe 🙏
As a matter of fact sir , I have solved this problem during my JEE days and have come to IIT BHU after qualifying JEE23 . Thank you sir for bringing out such problems
Nice Explanation Sir. but I have another approach for this question as we know that CONTINUED PRODUCT(k=1 to n-1) sin(kpi/n)=n/(2^(n-1) HERE S= sin(pi/7) x sin(2pi/7)x sin(3pi/7) S=sin(6pi/7)sin(5pi/7)sin(4pi/7) =>S^2 =CONTINUED PRODUCT(k=1 to 6) sin(kpi/7)=7/2^6 =>S= sqrt(7)/8 HOPE U LIKE IT!! THANK YOU
Your solution seems to be based on an interesting topic can you tell me more about that and for me there are some things that are unclear in your solution like how is s is equals to 2 products which are different I mean can you elaborate your solution
@@Player_is_I First, it's S^2 . Note that after squaring, each terms on RHS appear twice so that now we have a pair of each term which was present in S. Now , take exactly one term of each pair and use the identity sin(x) = sin(π-x) to get the desired product.
@@wipergaming1448 dekho anuj yahan sin p/7 sin2p/7 sin3p/7 ki bat hai aur ye teeno non zero hain aur teen non zero quantity ka product bhi non zero hi hoga
Good explanation sir 👏🏻.. but we can also solve this by following method Cosπ/7.Cos2π/7.Cos(π-4π/7) = Cosπ/7.cos2π/7.cos4π/7 =Sun(2.4π/7)/2³.sin(π/7) =Sin(π+π/7)/8sinπ/7 =Sinπ/7/8sinπ/7 =1/8 Thankyou!!!!
There's one more method ∏^n−1k=1sin(kπ/n)=n/[2^(n−1)] we need to find sinpi/7.sin2pi/7.sin3pi/7 but to use the formula Given above we need to have 6 terms as k goes from 1 to 6 we also know that sin(pi-pi/7)=sin(6pi/7), sin(pi-2pi/7)=sin(5pi/7), sin(pi-3pi/7)=4pi/7 so we can say that [sin(pi/7).sin(2pi/7).sin(3pi/7)]^2 = ∏^6 k=1sin(kπ/7)=7/[2^(6)] so now we can take root on both the sides and get our answer as root 7/ 8
@@slimshadyff7848 explain what? We only need to use the formula for this method, but the problem is most of us aren't aware of this formula. To calculate answer using the formula we need to make sure that we've 6 terms as the formula requires us to use from k=1 to k=(n-1) here n is 7 so k will be 6. Rest is basic, sin(pi-x) = sin x as second quadrant sinx is positive
sir have very fast solution for this with complex no there is a trig identity of sin(pi/n) *sin(2pi/n) .... *sin((n-1)pi/n) = n/2^n-1 now keep n = 7 and now sin 4pi/7 is same as 3pi/7 and so on therefore ans = sqroot(n/2^n-1)
Sir there is a shortcut to it like a formula is there where cos(a)*cos(2a)*cos(4a)... Where angles are in gp with common ratio 2= sin(twice of last angle)/2* no of terms*sin(first angle) and from there it's just a single liner question
Sir this question can be easily solved Let y = π/7 x = sin y × sin 2y × siny 3y x = (sin y × sin 2y × sin 3y × sin 4y × sin 5y × sin 6y)/ x [ because sin 4y = sin 3y , sin 5y = sin 2y and sin 6y = sin y] By using Π( multiplication of series) sin rπ/N = N /2^(N-1) x = 7/[(2^6) × x] x² = 7/2^6 x = √7 / 2³ = √7 / 8 Sir please reply to my method
I have created formula see here for n=7 x=√7/8=} x^2 = 7/(8^2) And for n x^2=n/(2^[n-1]) ...😊 Edit:- By the way 10 month pahele create Kiya tha abhi toh IITG 🥰 MAI baitha hu .❤
This formula is wrong.. it works only for this question.. For example, Try using it with sinπ/8, sin3π/8, sin5π/8 and sin7π/8 The value by formula comes as 1/4 But the actual value is 1/8
If I get this question, thsn I will focus on other problems than this one. IIT JEE failed person (rank 2k in 2004) this side..lol Probably justifies why I didn't qualify for IIT but doing better than 50% IITians by knowing how the world money flows.
guruji ; ye formula tabhi lgta ha jab cos ke angles gp me ho or common ratio 2 ho ; jese cos pi/7* cos 2pi/7*cos 4pi/7*cos 8pi/7. har jaagah cos ke multiple me ye formula nhi pel skte
apart from hard this seems to be quite a pointless problem in trigonometry, as both engineering and scientific point of view it has no significance, just few stupid formulas to learn to screw the lives of students and solve a pointless problem, how on earth would any student will logically come to the next step of getting the answer, this is mug up the solution kind of problem
Sir you can solve using π/7 property 4π/7=π -3π/7 It will give ans in 2 steps Sinπ/7 sin2π/7 sin4π/7 multiply divide by 2cosπ/7 and it will give ans 1/8
Hum is problem ko 2 to 3 lines mein bhi solve kar sakthe hein.. Sin pie/7 sin 2pie/7 sin 3pie/7 ko square kare tho sin(2n pie/2n+1) ka format ayega tho uska answer ayega 7/64 tho uska square root lelenge tho Aya root7/8
This level of complex math is obviously not made for the average person. Its required to get into IITs and other colleges depending on the performance, the better you do, the better your placement, that's the system, there is NO application for it in your daily life.
I learnt trigonometry in the early 70s.Now a retired banker. You said that the problem is with the approach, and not with formulae. Right? My question is why does my approach gets to be wrong. It was incorrect 50 years back and is incorrect even now.
I am read in class 6 th but I love seening this video l can understand this video 😊😊 and I am not joking😊mane apne maths wale sir se ye swal pucha tha but unohe ne 22÷7 man liya fir pura galt kar diya 😊 and i am preparing for Jee advanced and my some friend jeolus me and ma kabhi bhi 1st nhi aaya hu and i am watching this video 2024 😂 and mera result 5 aprail ko aayenga 😊❤and hare Krishna
x aur y me jo sin aur cos ke values diye hai na usme cos ke aur sin ke angles same hai toh hame ye a,b,c vale chakkar me fasne ki jarurat hi nhi hai let y= cos(p)=1/8 then x=sinp=√(8²-1²)/8=√7/8 Sir pin kardo dil khush hogaya hoga aapka ye mera comment padhke
Sir sin(2pie/7)=sin(5pie/7) As we take 7@=(2n-1)pie We can make sin4@=sin3@ Expanded to °4cos@(1-2sin²@)=3-4sin²@(because sin@=0 is rejected) °then squaring both sides 16(1-sin²@)(1-2sin²@)²=(3-4sin²@)² °-64sin⁶@+128sin⁴@-80sin²@=9+16sin⁴@-24sin²@ °64sin⁶@-112sin⁴@+56sin²@-7=0 (is the equation in which if we take roots in form of sin²@, roots are sin²(pie/7),sin²(3pie/7) and sin²(5pie/7)) Lets comment on the values for all three angle sin give positive value ) From equation produts of roots in form of sin²@: sin²(pie/7)sin²(3pie/7)sin²(5pie/7)=7/64 Then sin(pie/7)sin(3pie/7)sin(5pie/7)=square root (7) /8 Sir I did this please see this one time sir .
This question deserves to be the most hated problem of trigonometry
Phir sir ne kya bola wahi to bola hai 🙄
4 step mein hogaya 😂
Naa!!
@@shoyo_Ishida solution chaiye
@@aakritkatyal9135 noop!!
Long but predictable method: sin (kpi/m), k=1..n are roots of equation sin(mx) = 0. So we expand sin(mx) in powers of sin(x) to get a polynomial equation in sin(x) of degree m. For m = 7 we get:
sin(7x) = -64sin^7(x) +112 sin^4(x) -56 sin^2(x) + 7sin(x) =0. We can eliminate sin(x)=0 as solution, so the products of the remaining 6 roots is 7/64. This product is square of the required quantity.
Did you "predict" the method prior to looking at the solution?
Yeah bro same 😂 I once was curious and derived sin5x , 7x 9x 11x and those r some neet results if u want an interesting question check this .. what is the coefficient of last 3rd term of sin101x given that it's expansion is organised on the basis of descending order of powers of sinx ... Sayd question smjh me aagya hoga kya bol rha ....
Actually mein general term ka coefficient khoj rha tha nahi mila but kuchh terms me pattern mila
Hey please tell which theorem or Formula you have used, please tell the name 😢
We can use complex numbers where w = cos(2pi/n) where w is nth root of unity
very well explained and nice approach. thank you!
Alternate method is by complex No
z⁷ = 1 ( Roots are 1 , x,x².... x⁶)
x and x⁶ , x² and x⁴ ,x³ and x⁵ are conjugate
Now
z⁷-1 =(z-1)(1 -x)(1-x²)...(1-x⁶)
Collect the above pairs that are conjugate and simplify
z⁷ -1=(z-1)( z² - 2Re(x) +1)..(z² - 2Re(x³) +1 )
Put the values in cos form of Re(x) and put z=2
Sir, i saw your video on the same problem and the next morning when i was searching for the video, i didn't find it anywhere, not even in my history, but then i thought that you might have deleted the video. Thanks for re uploading the problem.
Same with me😅
"Maths approach ka game hai " 🔥🔥
भाई मुझे तो पता ही नहीं था l
Thanks for telling .
In general , it can be proved using complex numbers that: multiplication of 'N' terms of sines where each acute angle is in A.P. such that the first term is equal to common difference and last term is just less than π , is equal to (N+1))/2^(N).
In the question in this video , let the value to be determined be X . Then , X²= sin(π/7)sin(π/7)sin(2π/7)sin(2π/7)sin(3π/7)sin(3π/7)
Now using sin(x) = sin(π-x) , we have ,
X^2 = sin(π/7)sin(2π/7)sin(3π/7)sin(4π/7)sin(5π/7)sin(6π/7)
The last equation follows the formula above , hence , X^2 = 7/2^6 = 7/64 . Take square root and you get your answer.
What should I put in the value of N
@@vansh61595 As written above , N is the number of terms.
Acha , thik h
Hello :)
I wanted to know how you came to know about this identity, because I am halfway through my 11th std., starting the chapter 'Circles' tomorrow, and I never came across such an identity while studying the two trigonometry chapters in the past. Thanks.
Thank you for this trick
Sir this question could be done easily by using the concept nth root of unity,
Where n is 7.
(x^7-1)/(x-1)=1+x+x^2+...+x^6=(x-e^i2pie/7)....(x-e^i12pie/7)
Club conjugate vise all the 6 factors in rhs.
They will become 3 ,
(x^2 - 2xcos(2pie/7) +1)...
Put x=1
Lhs = 7
Rhs = (8)(1-cos(2pie/7))....
And rhs = 64(sin(pie/7).sin(2pie/7).sin(3pie/7))^2
So got it
Well explained sir!! I appreciate your effort 😊👍 But sir ,,i have an another excellent approach,,which is luckily a brilliant and time saver approach for this particular question!!
We have a formula ,,,sinπ/n .sin 2π/n .sin3π/n ...
.sin (n-1)π/n = n/{2^(n-1)}.
In this particular problem we have value for sinπ/7 sin2 π/7
...sin6π/7 = 7/(2^6)..
Now we let us assume sin π/7...sin3π/7 =X .then the next three terms can be transformed in terms of Series of X.as eg:-sin4π/7 =sin 3π/7..now the series become X^2 in LHS...now apply root both side .we get value as √7/8..
Its so simple,,can be done within 4-5 lines😊😎🔥🤘 Lets have a look to my approach! Hope u like it !!.
Thankyou!
Radhe radhe🙏
Could you pls share with me a link for the proof of this formula, or just hint me the approach to prove it? Thanks
@@shauryagupta3644i appreciate your curiosity 😊!! And thankyou for read my approach 😊 .
To be honest ,,this formula is not related to intermediate level directly!! And the proof is so lengthy ,you have to use eulers identity ,concept of limit ,and little bit use of complex no. Yeah iam also an iit jee aspirant right now😊 so ,no doubt you can understand the whole proof almost ! But this will worthless for u right now !! The result is important ,,so plz try to keep that result in mind ! Remember,,you cant use this formula in every situation,,the major condition is the coefficient of π in last term is 1 less than the denominator !!and the product should be continuous!! Further modification in this formula led to the result of "Riemann integral ". But still you want to know the proof ,,i have done an effort in searching the video you need to get the proof . Iam providing you link ,,you can chek it out!😊👇
ruclips.net/video/hQ532SjbxNo/видео.htmlsi=auUbzdDu8AipVGw_
This contain the proof u want,,and given by any university professor !!👍.
Once again ,i say ,,result is important!! Thankyou to love my approach!!😊👍
Radhe radhe 🙏
@@shyamaldevdarshan thanks I applied without knowing proof definately gonna learn proof after my complex number course.
@@timewalkwalker welcome! Yeah sure! You can learn😊👍
Radhe radhe🙏
Yes
As a matter of fact sir , I have solved this problem during my JEE days and have come to IIT BHU after qualifying JEE23 . Thank you sir for bringing out such problems
Hey did you solve this by this method only ?
Sale flex maar. Rha h tu
@@itsshubh3430 If he really did this tough problem all on his own he can flex and deserves the honour,kudos to the guy👏👏
Congrats bro❤
Q@@😊@😊
When we got 8xy=x, why can't we do this to find x 👇
8xy-x=0
x(8y-1)=0
x=0, y=1/8
It would've been much faster
wrong if y=1/8 x can be any value
@@adarshsubramanian9560 bro xy = x/8 wat are you talking about
@@yxsh5033 For example if y=1/8 then if u take x value as 1 then also equation is satisfied
because x is not equal to zero
Maza aa gaya sir,
Definitely looking up for more such videos❤
Nice Explanation Sir.
but I have another approach for this question
as we know that CONTINUED PRODUCT(k=1 to n-1) sin(kpi/n)=n/(2^(n-1)
HERE S= sin(pi/7) x sin(2pi/7)x sin(3pi/7)
S=sin(6pi/7)sin(5pi/7)sin(4pi/7)
=>S^2 =CONTINUED PRODUCT(k=1 to 6) sin(kpi/7)=7/2^6
=>S= sqrt(7)/8
HOPE U LIKE IT!!
THANK YOU
Yes. It's a general formula which can be easily proved using complex numbers and a little bit algebraic manipulations.
This is exactly what i thought when I saw this.
Can we get in touch please I think u can help me in mathematics.
Your solution seems to be based on an interesting topic can you tell me more about that and for me there are some things that are unclear in your solution like how is s is equals to 2 products which are different I mean can you elaborate your solution
@@Player_is_I First, it's S^2 . Note that after squaring, each terms on RHS appear twice so that now we have a pair of each term which was present in S. Now , take exactly one term of each pair and use the identity sin(x) = sin(π-x) to get the desired product.
5:20 x zero nahi ho sakhta aapne kaise judge kiya sir? Plz batadijiye.
Kyunki sin. 0 180 360 me hi zero hota hai baki jagah nahi
@@neerajprajapati3852 bhai lekin x toh multiplication of sin fn hai toh vo zero bhi ho sakhte hai.
@@wipergaming1448 dekho anuj yahan sin p/7 sin2p/7 sin3p/7 ki bat hai aur ye teeno non zero hain aur teen non zero quantity ka product bhi non zero hi hoga
Kaisa hai @@wipergaming1448
Jab bhi ham kisiko bhi cancel karenge ek probability hothe hai ki vo zero ho sakta hai
Good explanation sir 👏🏻.. but we can also solve this by following method
Cosπ/7.Cos2π/7.Cos(π-4π/7)
= Cosπ/7.cos2π/7.cos4π/7
=Sun(2.4π/7)/2³.sin(π/7)
=Sin(π+π/7)/8sinπ/7
=Sinπ/7/8sinπ/7
=1/8
Thankyou!!!!
Buddy don't u think ki tumhare method se answer -1/8 aaega because cos(180 - x ) is -cos(x) 🤔
There's one more method
∏^n−1k=1sin(kπ/n)=n/[2^(n−1)]
we need to find sinpi/7.sin2pi/7.sin3pi/7
but to use the formula Given above we need to have 6 terms as k goes from 1 to 6
we also know that sin(pi-pi/7)=sin(6pi/7), sin(pi-2pi/7)=sin(5pi/7), sin(pi-3pi/7)=4pi/7 so
we can say that [sin(pi/7).sin(2pi/7).sin(3pi/7)]^2 = ∏^6 k=1sin(kπ/7)=7/[2^(6)]
so now we can take root on both the sides and get our answer as root 7/ 8
Awesome ❤
Ans the proof of that formulae comes from complex numbers
Can you please explain that to me 😅
@@slimshadyff7848 explain what? We only need to use the formula for this method, but the problem is most of us aren't aware of this formula. To calculate answer using the formula we need to make sure that we've 6 terms as the formula requires us to use from k=1 to k=(n-1) here n is 7 so k will be 6. Rest is basic, sin(pi-x) = sin x as second quadrant sinx is positive
@@hretzbro pls can you give name of the formula so that i can learn it .
"BHANNAAAAT IS A ICONIC WORD"
LOVE YOU SIR❤❤❤
sir have very fast solution for this with complex no there is a trig identity of sin(pi/n) *sin(2pi/n) .... *sin((n-1)pi/n) = n/2^n-1
now keep n = 7 and now sin 4pi/7 is same as 3pi/7 and so on therefore
ans = sqroot(n/2^n-1)
Appriciate for those student who solved. This question 😮
Bruh it's a basic question 🤖
@@MehtabHussain-vh3bljee also ask basic question
why video was deleted and reuploaded sir?
Very nice sir. Quite inspirational solution
Great question sir ,thank you sir for providing this type of questions.
Sir please derive formula . I always mug up and can't understand relationship between the formula . √( s . S- a . S- b. S- c)
Just search for derivation of Heron’s formula
It's easy you can find so many videos on it in English and may be in Hindi too
Req to you sir..pls start PYQ solving of last 3-4 years Maths
Good qustion sir jeee namste🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
Super explanation sir 👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻🥳🥳🥳🥳
You are really a great teacher who can turn the most hated question to most loveliest one.
#remember as result(RAR)=1)approach 2)result
Reupload?
Sir there is a shortcut to it like a formula is there where cos(a)*cos(2a)*cos(4a)... Where angles are in gp with common ratio 2= sin(twice of last angle)/2* no of terms*sin(first angle) and from there it's just a single liner question
usse ye wala nahi ban rha kyuki gp nahi hai
Sir this question can be easily solved
Let y = π/7
x = sin y × sin 2y × siny 3y
x = (sin y × sin 2y × sin 3y × sin 4y × sin 5y × sin 6y)/ x [ because sin 4y = sin 3y , sin 5y = sin 2y and sin 6y = sin y]
By using Π( multiplication of series) sin rπ/N = N /2^(N-1)
x = 7/[(2^6) × x]
x² = 7/2^6
x = √7 / 2³ = √7 / 8
Sir please reply to my method
Second equation will be x2
Good evening sir.Sir pyq solving videos please
I have created formula see here for n=7 x=√7/8=} x^2 = 7/(8^2)
And for n x^2=n/(2^[n-1]) ...😊
Edit:- By the way 10 month pahele create Kiya tha abhi toh IITG 🥰 MAI baitha hu .❤
Yeah easiest method
how did u derive this?
This formula is wrong.. it works only for this question..
For example,
Try using it with sinπ/8, sin3π/8, sin5π/8 and sin7π/8
The value by formula comes as 1/4
But the actual value is 1/8
@@NeelGunjal01Your expression is different see for n=8 it should be sin(π/8) , sin(2π/8) , sin(3π/8), sin(4π/8) and it will true...
@@cofanavay2235 How do you determine what the expression must be?
Complex numbers se kijiye alag swaad aayega. (use 7th root of unity)
Bro kaise
Kaise Bhai thoda hint batao
Ye questions agar adv mein kiya toh mera IIT jayengaa....😂
As this is a IMO problem.............So This should be one of the hardest problem....
If I get this question, thsn I will focus on other problems than this one. IIT JEE failed person (rank 2k in 2004) this side..lol
Probably justifies why I didn't qualify for IIT but doing better than 50% IITians by knowing how the world money flows.
So how exactly does money flow
Solved it...!... Loved the problem... I tried straight for two days for it 😂
Can u please solve this question using quadratic equations/polynomials?
Wow!
nobody will mind even if you also declare it as the most hated question in JEE level.
Can also be done by complex no
Direct formula for sin(pi/n).sin(2pi/n)….sin((n-1)pi/n) = n/(2^(n-1))
Derived in complex numbers
Last me root bhi aayega n
Bekar
Sin ko cos me covert karo aur
CosA cos2A cos4A....= sin2n/2nsina
guruji ; ye formula tabhi lgta ha jab cos ke angles gp me ho or common ratio 2 ho ; jese cos pi/7* cos 2pi/7*cos 4pi/7*cos 8pi/7. har jaagah cos ke multiple me ye formula nhi pel skte
Different approach - for series that is in product and is in form of sin theta we can use √2n+1/ 2^n . N - no of terms
Use of complex no
Abe chup gawar
It's just a easy question of trigonometric series
As a 12th class student i can guarantee guys this is nothing . If u are an engineering student u have to been through a lot .
I can guarantee you could not have solved this problem had you not seen the solution in advance.
So this problem is SOMETHING.
apart from hard this seems to be quite a pointless problem in trigonometry, as both engineering and scientific point of view it has no significance, just few stupid formulas to learn to screw the lives of students and solve a pointless problem, how on earth would any student will logically come to the next step of getting the answer, this is mug up the solution kind of problem
अतिसुंदर❤
Sahi bola sir, formula hai sara yaad pr question nhi bn pata mere yaar
Sir you can solve using π/7 property 4π/7=π -3π/7
It will give ans in 2 steps
Sinπ/7 sin2π/7 sin4π/7 multiply divide by 2cosπ/7 and it will give ans 1/8
I am also getting same
Sin m^n=? ( No expansion, should be FINITE SERIES.)
Sir apse defense ka preparation krna hai kaise kre ❤
why are we not considering the negetive counterpart of x?
sir!! PW kyu chhoda ??
Sir wat is math?
Kya sir apply formula of trigonmetric series😮
Bro there is only formula for continued product of cosine series and not of Sine series
sare sin ke angle pi by 2 karke cosine banake bhi ho sakta hai
Hum is problem ko 2 to 3 lines mein bhi solve kar sakthe hein..
Sin pie/7 sin 2pie/7 sin 3pie/7 ko square kare tho sin(2n pie/2n+1) ka format ayega tho uska answer ayega 7/64 tho uska square root lelenge tho Aya root7/8
Actually, this question is from a book named A das gupta
The basic orthogonalality of sine waves is demonstrated here.
My whole high school life ended learning this sin-π cos-π but still i dont know where to use this math formulas in my daily life.
This level of complex math is obviously not made for the average person. Its required to get into IITs and other colleges depending on the performance, the better you do, the better your placement, that's the system, there is NO application for it in your daily life.
Re-upload 😊
Toh last time the answer was wrong as he forgot to multiply by 8
We love you sir❤
5[1.96-sin^2{(cos^-1 (0.02)}° ] find the value of it . Challenge to you sir 😊😊😊.
I am read in class 9
Use continuous product of sine
Direct solve ho Jayega easy hai
Loved it ❤
Sir kya ye question class 10 ke liye bhi hai
Master ji jab y= ......... Jab diya hi nahi hai to apne kaise man liya
2 ko 4 time multiply karne se 8 kaise hoga sir.
Kya sir ne yeh video doobara post kra hai isse pehle yeh video sir ne kya post Kiya tha jo delete kr diya tha koi batao
Sir ... is it the only solution for this question..... if any solution is there please tell
Apply direct formula for product series
Check my comment 😊👍
@@shaatirop8901WHAT??WHICH FORMULA
Sir apke questions ❤❤❤❤
Mujhe bhi apne EX ki talash hai, chhodunga naheen bas ek baar mil jaye...
Sir please NDA ka batch start kr do yt pe please sir😢😢😢😢🙏🏼🙏🏼🙏🏼🙏🏼🙏🏼🙏🏼
Need aa gyi dekhte dekhte😮
Excellent sir
Thanks
I learnt trigonometry in the early 70s.Now a retired banker.
You said that the problem is with the approach, and not with formulae. Right?
My question is why does my approach gets to be wrong.
It was incorrect 50 years back and is incorrect even now.
sir 3 step ma aap ye jo tod te ho ye samaj ma nahe aata ha jesa ke aao ne 4π/7 ko 3π/7 likh diya 😢
You should also prove why you took the positive part
There is a formula for this question followed by multiplying dividing it by 2
Legends ->> repload
Bhai ye sin theta kya hota hai? 😢
Konsi class mai h bhaii?
@@Ridham.khatri bhai ye sab cheesein mera jeena haram kar diya 12th class mein. Sapne mein bhi ye theta veta dekhke mein dar se mar jaoonga
Arjuna 1.0 adv lecture tarun sir ne iske 3 methods bataye the 7 months pehle 😂
Pr sir na ak method se solve krdiya it's talent😂
Mt sir ne 10 Saal phele💀💀
@@Devansh_6655 mt sir ka lecture?
Very easy question of complex no.
I am read in class 6 th but I love seening this video l can understand this video 😊😊 and I am not joking😊mane apne maths wale sir se ye swal pucha tha but unohe ne 22÷7 man liya fir pura galt kar diya 😊 and i am preparing for Jee advanced and my some friend jeolus me and ma kabhi bhi 1st nhi aaya hu and i am watching this video 2024 😂 and mera result 5 aprail ko aayenga 😊❤and hare Krishna
Radhe Radhe hare Krishna ❤❤❤
Vo wahar lal nehru scool❤
Lol
😂😂😂😂 learn counting from 1 to 100 properly.
Edit. Thanks for 2k likes.
Nice joke@@MaheshKumar-lx1ku
Let me do it
Sin 22/7 /7 will take common then(2*3)
Sin22*6
Sin132 😅
Maza aagaya sir but itna time kidhar hai solve karne ke liye ek question ko
legendary, 2 saal purana doubt solve hogaya
Sir wbjee k liye kuch class hota to acha hota
Khatarnak question
Nice sir❤
it can be easily done by complex number tried and tested
Sir ne bhot bade tarike se karvaya hai
Sir isko complex no se kiya Jaye to 2 min me ho gayegaa
You are the best guruji, thanks for the question practices ❤❤
Can also be done by complex no.
Wonderful !
x aur y me jo sin aur cos ke values diye hai na usme cos ke aur sin ke angles same hai toh hame ye a,b,c vale chakkar me fasne ki jarurat hi nhi hai let y= cos(p)=1/8 then x=sinp=√(8²-1²)/8=√7/8
Sir pin kardo dil khush hogaya hoga aapka ye mera comment padhke
Sir sin(2pie/7)=sin(5pie/7)
As we take 7@=(2n-1)pie
We can make sin4@=sin3@
Expanded to
°4cos@(1-2sin²@)=3-4sin²@(because sin@=0 is rejected)
°then squaring both sides
16(1-sin²@)(1-2sin²@)²=(3-4sin²@)²
°-64sin⁶@+128sin⁴@-80sin²@=9+16sin⁴@-24sin²@
°64sin⁶@-112sin⁴@+56sin²@-7=0 (is the equation in which if we take roots in form of sin²@, roots are sin²(pie/7),sin²(3pie/7) and sin²(5pie/7))
Lets comment on the values for all three angle sin give positive value )
From equation produts of roots in form of sin²@:
sin²(pie/7)sin²(3pie/7)sin²(5pie/7)=7/64
Then
sin(pie/7)sin(3pie/7)sin(5pie/7)=square root (7) /8
Sir I did this please see this one time sir .
Assumption ka game hai...