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
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!😊👇 ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-hQ532SjbxNo.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.
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
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 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
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
@@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
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
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.
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 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 .
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
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
As a student of 10th class, I think trigonometry is used more in making questions harder , instead of finding length of a building.😂😂😂 Edit:Oh my god mai toh famous ho gya.
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.
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
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
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.
mujhe to sapna aayega ki kaun sa formula lagana hai aur kuansa nhi .... Sir ko aa gaya .... ham phase rhe...😂 but ab video dekhne ke baad samjh gaya hoon...thank you
Let T=sin(pi/7)sin(2pi/7)sin(3pi/7) T=sin(6pi/7)sin(5pi/7)sin(4pi/7) Multiply both equations T^2=sinpi/7)....sin(6pi/7) Then put formula sin(pi/n)sin(2pi/n)....sin((n-1)pi/n)= n/2^{n-1)
Don't do this long bro, we know that sin(π/n)sin(2π/n)....sin[(n-1)π/n]=n/[2^(n-1)] In this question put n=7, multiply divide by sin(π/7)sin(2π/7)sin(3π/7) and assume it T T = (7/2^6)/T T=√7/8 👍🏻 Easy
