Asimov Pupil Posted April 11, 2005 Share Posted April 11, 2005 8((sin2a)^2)-4 as a single function of sin(a) or cos(a) sry Chapter final i had no idea about and this was the only one that slipped me. Link to comment Share on other sites More sharing options...
Asimov Pupil Posted April 11, 2005 Author Share Posted April 11, 2005 by the way anyone know how to put say y^2 so that it looks like y squared? Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted April 11, 2005 Share Posted April 11, 2005 y2 That's using [ sup] tags (without the space). Or y² alt 178 Link to comment Share on other sites More sharing options...
dan19_83 Posted April 11, 2005 Share Posted April 11, 2005 Using the formula that Sin2(a) = 2Sin(a)Cos(a) Bring the 2 outside the bracket and multiply by the 8: 16[(Sin(a)Cos(a))^2]-4 Then you multiply the Sin(a)Cos(a) by itself and this changes the formula to: 16[sin^2(a) + 2Sin(a)Cos(a) + Cos^2(a)]-4 Since Sin^2(a) = 1- Cos^2(a) The formula changes to: 16[1- Cos^2(a) + 2Sin(a)Cos(a) + Cos^2(a)]-4 The two Cos^2(a) cancels leaving: 16[1 - 2Sin(a)Cos(a)]-4 Multply out and you get: 16 - 32Sin(a)Cos(a) -4 Leaving you with an answer of: 12 - 32Sin(a)Cos(a) I hope this makes sense to you because as you may have figured out, i don't know how to put in ^2, which would make it easier to read. Anyway if you don't understand it, just say the word and i will happily try and re-do it in a better format. Link to comment Share on other sites More sharing options...
Asimov Pupil Posted April 11, 2005 Author Share Posted April 11, 2005 hmm i get that and thank you, but how do you put it as just a sin(a) function THank you Link to comment Share on other sites More sharing options...
dan19_83 Posted April 11, 2005 Share Posted April 11, 2005 oops, thought it was sina and cosa! i'll get back to you Link to comment Share on other sites More sharing options...
dan19_83 Posted April 11, 2005 Share Posted April 11, 2005 2Sin(a)Cos(a) = Sin^2(a) So the formula at the end becomes: 16 - 16(Sin^2(a)) How's that? If you want to put it in terms of cos then: 16(1-Sin^2(a)) = 16 Cos^2(a) Link to comment Share on other sites More sharing options...
Asimov Pupil Posted April 12, 2005 Author Share Posted April 12, 2005 2Sin(a)Cos(a) = Sin^2(a) So the formula at the end becomes: 16 - 16(Sin^2(a)) How's that? If you want to put it in terms of cos then: 16(1-Sin^2(a)) = 16 Cos^2(a) hey i thought it was 2sin(a)cos(a)= Sin2(a) Link to comment Share on other sites More sharing options...
Dave Posted April 12, 2005 Share Posted April 12, 2005 Indeed. [math]2\sin(a)\cos(a) = \sin(2a)[/math]. Substitute this directly into your formula and use the fact that [math]\cos^2(x) = 1 - \sin^2(x)[/math]. Link to comment Share on other sites More sharing options...
dan19_83 Posted April 12, 2005 Share Posted April 12, 2005 hey i thought it was 2sin(a)cos(a)= Sin2(a) You are correct. my mistake. must look into it a bit further so! Try and come up with some suggestions yourself if you can. Link to comment Share on other sites More sharing options...
dan19_83 Posted April 12, 2005 Share Posted April 12, 2005 Indeed. [math]2\sin(a)\cos(a) = \sin(2a)[/math]. Substitute this directly into your formula and use the fact that [math]\cos^2(x) = 1 - \sin^2(x)[/math']. That would work except that he is looking for it in terms of Sin(a) and not Sin (2a). I've really gotta learn how to put in those fancy formulas! Link to comment Share on other sites More sharing options...
Dave Posted April 12, 2005 Share Posted April 12, 2005 That would work except that he is looking for it in terms of Sin(a) and not Sin (2a). The method I gave will give it in terms of sin(a). Link to comment Share on other sites More sharing options...
dan19_83 Posted April 12, 2005 Share Posted April 12, 2005 But if you sub in your sin(2a) into the formula I worked out then you would get: 16-16Sin(2a). Anyway it's too late for this (3a.m.!!), I'm going to bed! Link to comment Share on other sites More sharing options...
Asimov Pupil Posted April 12, 2005 Author Share Posted April 12, 2005 8((sin2a)^2)-4 8(sin2(a))(sin2(a))-4 8(2sin(a)cos(a))(2sin(a)cos(a))-4 32(sin^2(a))(cos^2(a))-4 32sin^2(a)(1-sin^2(a))-4 32sin^2(a)-32sin^4(a)-4 4(8sin^2(a)-8sin^4(a)-1) now what pls ps. just a sin(a) or cos(a) funtion not to any power Link to comment Share on other sites More sharing options...
Dave Posted April 12, 2005 Share Posted April 12, 2005 You can try factoring it, but you're probably not going to remove the sin2. I'm afraid I don't have time to look at this problem further atm; I shall look at it in a few hours time. Link to comment Share on other sites More sharing options...
Asimov Pupil Posted April 12, 2005 Author Share Posted April 12, 2005 thank you Link to comment Share on other sites More sharing options...
Asimov Pupil Posted April 12, 2005 Author Share Posted April 12, 2005 BAH! it's 4cos4a (take out a negative four, then use pythagorean therom!) thanks for the help by the way Link to comment Share on other sites More sharing options...
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