Patel108 Posted January 19, 2011 Report Share Posted January 19, 2011 Do you guys know of any websites or other free materials where I can learn this stuff? as in What is the sum of the infinite series 1 + sin x + (sin x)^2 etc......or prove that cos x + cos 3x + cos 5x + ......... + cos (2n-1)x = (sin 2nx) / 2 sin x ----- stuff like thatOne more thing that I can't seem to figure out:How does sin2x(2cos2x) = sin4x ? Thanks. Reply Link to post Share on other sites More sharing options...
dessskris Posted January 19, 2011 Report Share Posted January 19, 2011 Your textbook...?Sum of such series can be found by finding the f(x) when x=1, 2, 3, 4, 5, 6, etc..Then if you see such pattern, find the general formula and that's your answer. Or you may try to plot the data and find the equation of the function.Are you asking help to prove that cos((2n-1)x) thing or not? Prove it by induction, right? If you want I can try to do it and if I can get the answer, I can scan my working and upload it here sin(2x) * 2cos(2x) = sin(4x)sin(2x) * 2cos(2x) = 2 * sin(2x) * cos(2x)which is true...?Remember the double angle rule?sin(a+b) = sin(a) * cos(b) + cos(a) * sin(b)And in that case, a=b=2x Reply Link to post Share on other sites More sharing options...
Patel108 Posted January 20, 2011 Author Report Share Posted January 20, 2011 My textbook doesn't have this stuff at all.........No I am not asking for you to solve it, but if there are any other resources that show proof by induction using trigonometry. Although I do have another question (sorry if this seems so random).How does 2*sin((2^k)x)*cos((2^k)x) = sin((2^(k+1))x) ? Reply Link to post Share on other sites More sharing options...
dessskris Posted January 20, 2011 Report Share Posted January 20, 2011 Woah really? What textbook are you using? Ask your teacher to get another textbook and photocopy it for your class! Hmm you may try google lah..or go to the Links section in this forum and see the Math links.I got the proof for you below btw. And also the formula. My handwriting is ugly btw so if you can't read it please let me know. 3 Reply Link to post Share on other sites More sharing options...
Patel108 Posted January 21, 2011 Author Report Share Posted January 21, 2011 Thank you very much for the proofs!!!!!Yeah, our class is using the Cambridge Mathematics for the IB Diploma Higher Level 1 Book by Hugh Neill and Douglas Quadling. Most of the stuff that she is teaching us (well actually she doesn't teach that much) is not found in our textbook in all. It's in the Higher Level 2 textbooks, so if we don't really understand anything, then we're screwed .I tried looking it up on google and the Links tab, but I still couldn't find any helpful resources. Reply Link to post Share on other sites More sharing options...
dessskris Posted January 21, 2011 Report Share Posted January 21, 2011 Eh I just realised that the second picture did not appear. Did you see two pics? Only 1 right? Thank God I still have the hard copy just scanned and reuploaded it. I hope it will work this time.I have the same textbook, both books 1 and 2. My teacher seldom uses it, though. She makes us notes and explains stuffs in her own way so the textbooks are kind of useless and they are so heavy!But, if you have anything, just come here and we will try to help you! Reply Link to post Share on other sites More sharing options...
nik_20 Posted June 14, 2011 Report Share Posted June 14, 2011 Ok, so I have the exact same question....lolBut how would you show that cos x + cos 3x + cos 5x +...+ cos (2n-1)x = sin 2nx/2 sin x when n=2 and n=3?I would really appreciate your help!!! Reply Link to post Share on other sites More sharing options...
puyol9 Posted June 16, 2011 Report Share Posted June 16, 2011 (edited) She used mathematical induction to prove the above equation, that is, essentially she proved for all n (despite not showing that n=2,3 is true).If her proof is confusing, you NEED to go over mathematical induction (or learn it if you haven't already), ask your maths teacher about it, as it can be a little confusing to grasp the logic behind this type of proof. I am assuming you are doing the HL maths course, in which case you will need to know this method of proof. I am not sure whether SL students need to know this. If you want to verify for n=2 and n=3 (despite Desy Glau proves for all n):n=2:Proposition: cos x + cos3x = sin4x /2sinxLHS: cosx + cos3x=cosx + cosxcos2x - sinxsin2x (compound angle formula)=cosx + cosx(cos2x -sin2x) - sinx(2cosxsinx) (compound angle formulas)= cosx(1+cos2x -sin2 - 2sin2x)=cosx(2cos2x -2sin2x) (Pythagorean identity)= 2cosx(cos2x) (double angle formula)=2(2cosx*sinx)*cos2x/(2sinx)=2sin2x*cos2x/(2sinx) (double angle formula)=sin4x/(2sinx) (double angle formula)Therefore, LHS = RHSThus, n=2 is verifiedn=3 is done in a similar (but longer way) you will need to expand all cos(nx) then group and simplify, there may be a shorter proof for n=2, but this one is okay, but it definately gets longer as n increases (please note that I purposely didn't leave the proposition as an equation, instead I separated LHS and RHS, as this gives a more formal proof, not sure why but leaving it so that it has two sides and then manipulating both sides is frowned upon)PS. Not to be picky or anything Desy Glau, but you might want to be careful with your structure of your proof, as the IB examiners seem to be very PICKY about it, especially wording, and you will lose marks unless you have the exactly correct wording. Ie, instead of assume that when n=k you should say assuming Pk is true etc. Edited June 17, 2011 by puyol9 1 Reply Link to post Share on other sites More sharing options...
bomaha Posted June 17, 2011 Report Share Posted June 17, 2011 I think that this question was in a math HL past paper. Reply Link to post Share on other sites More sharing options...
puyol9 Posted June 17, 2011 Report Share Posted June 17, 2011 Do you guys know of any websites or other free materials where I can learn this stuff? as in What is the sum of the infinite series 1 + sin x + (sin x)^2 etc......or prove that cos x + cos 3x + cos 5x + ......... + cos (2n-1)x = (sin 2nx) / 2 sin x ----- stuff like thatFirst series:This is not a series. Sinx is a function, thus there is no one answer to the series 1 + sinx + sin2x + sin3xSecond:Desy Glau proved this above.Third:How does 2*sin((2^k)x)*cos((2^k)x) = sin((2^(k+1))x)?Using double angle formula:sin(2k+1x) = sin(2*2kx)As sin2a=2sina*cosa for any a (real number)If a = 2kxThen sin(2*2kx)=2sin(2kx)*cos(2kx)) Reply Link to post Share on other sites More sharing options...
sprintdominator Posted October 14, 2011 Report Share Posted October 14, 2011 LOL, To prove the summation series for trigo function, Apply de moivres theorem. !! Its sort of an application which no one else seems to know.The core of the solution is to recognise thatcos.x + cos.2x + cos.3x + ... + cos.nx= Real part of (cos.x + i.sin.x + cos.2x + i.sin.2x + cos.3x + i.sin.3x + ... + cos.nx + i.sin.nx)(where i = √-1)= Real part of [ (cos.x + i.sin.x) + (cos.x + i.sin.x)^2 + (cos.x + i.sin.x)^3 + ... + (cos.x + i.sin.x)^n ] ... [Eq'n 1]This is a geometric progression: let Y = (cos.x + i.sin.x), and the series becomesReal part of (Y + Y^2 + Y^3 + ... + Y^n)From the rules of a geometric progression, this is= Real part of [Y.(Y^n - 1) / (Y - 1)]Now substitute Y = (cos.x + i.sin.x) back in again, and the sum becomes= Real part of (cos.x + i.sin.x).[ (cos.x + i.sin.x)^n –1] / (cos.x + i.sin.x - 1)= Real part of (cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1] / (cos.x + i.sin.x - 1)Rationalise The DenominatorThe denominator is complex. To get rid of the “i” term, multiply numerator and denominator by the complex conjugate of the denominator, (cos.x –1 – i.sin.x), and the summation becomes=(cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1].(cos.x -1 + i.sin.x ) / [ (cos.x -1 + i.sin.x ).( cos.x -1 - i.sin.x) ]For clarity, the denominator will be expanded first:= (cos.x -1 + i.sin.x ).( cos.x -1 - i.sin.x)This is simply the difference of two squares:= (cos.x – 1)^2 – (i.sin.x)^2= (cos.x)^2 – 2.cos.x. + 1 – (i^2.(sin.x)^2)= (cos.x)^2 – 2.cos.x. + 1 + (sin.x)^2But (cos.x)^2 + (sin.x)^2 = 1, so the denominator becomes= 2 – 2.cos.x=2.(1 – cos.x)Evaluate the NumeratorThe numerator is= (cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1].( cos.x -1 - i.sin.x)= (cos.x.cos.nx + cos.x.i.sin.nx - cos.x + i.sin.x.cos.nx + i.sin.x.i.sin.nx - i.sin.x)).( cos.x -1 - i.sin.x )But from the standard trigonometric identities,(cos.x.cos.nx - sin.x.sin.nx) = cos(nx + x) = cos(n+1).x andcos.x.i.sin.nx + i.sin.x.cos.nx = sin(nx + x) = sin(n+1).xSo the numerator becomes= (cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x). ( cos.x - 1 - i.sin.x )= cos.x.(cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x) - (cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x) - i.sin.x.(cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x)= cos.x.cos(n+1).x + i.cos.x.sin(n+1).x - cos².x - i.cos.x.sin.x - cos(n+1).x - i.sin(n+1).x + cos.x + i.sin.x - i.sin.x.cos(n+1).x - i².sin.x.sin(n+1).x + i.sin.x.cos.x + i².sin.x.sin.x= cos.x.cos(n+1).x + i.cos.x.sin(n+1).x - cos².x - i.cos.x.sin.x - cos(n+1).x - i.sin(n+1).x + cos.x + i.sin.x - i.sin.x.cos(n+1).x + sin.x.sin(n+1).x + i.sin.x.cos.x - sin².x ... [Eq'n 2]= cos.x + cos.2x + ... cos.nx + i.(sin.x + sin.2x + ... + sin.nx)Find cos.x + cos.2x + cos.3x + ... + cos.nxFrom Equation [1], the sum of the series is the real part of the geometric progression:Gathering real terms, the real part of equation 2= cos.x.cos(n+1).x - cos².x - cos(n+1).x + cos.x + sin.x.sin(n+1).x - sin².xSimplifying,cos.x.cos(n+1).x + sin.x.sin(n+1).x = cos[(n+1).x - x] = cos.nx, and-cos²x - sin²x = -(cos²x + sin²x) = -1,so the real part becomes= cos.nx - cos.(n+1).x - 1 + cos.xSo the sum of cos.x + cos.2x + ... + cos.nx= numerator / denominator= (cos.nx - cos.(n+1).x - 1 + cos.x) / 2.(1 - cos.x)= (cos.nx - cos.(n+1).x) / (2 - 2.cos.x) - 1 / 2Q.E.D.To verify this, choose x = π / 5 and n = 3 as an example:(cos.3.π / 5 - cos.(4.π / 5)) / (2 - 2.cos.π / 5) - 1 / 2= 0.809cos(π / 5) + cos(2.π / 5) + cos(3.π / 5)= 0.809 + 0.309 - 0.309= 0.809 1 Reply Link to post Share on other sites More sharing options...
sprintdominator Posted October 14, 2011 Report Share Posted October 14, 2011 Hope this helps Unorthodox, but it works perfectly fine for summation series of trigo. Try deriving cos x + cos 3x + cos 5x + ......... + cos (2n-1)x = (sin 2nx) / 2 sin x using the above method. its not that difficult just that you have to take note thyat in this case, it is the sum of only the odd numbered terms. Try to, for summation series of sine and tangent!! haha Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.