Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hi
    Was wondering how you know which form to make it in?

    ie.
    looking at this question:

    2cos3theta - 3 sin3theta = -1

    would you put it in the form:

    Rsin (3theta + alpha) and go from there or
    Rcos(3theta + alpha)


    I have a sheet with an example Rsin(theta + alpha) which I've just been following for another question, it works, but I don't understand why. I hate this topic. -.-
    Offline

    0
    ReputationRep:
    I think you use the latter equation...so:

    2cos3theta - 3 sin3theta = Rcos(3theta + alpha)

    Expand the second equation, equate the co-efficients, cancel out the identical parts, then figure what R and alpha is from there.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Student29)
    I think you use the latter equation...so:

    2cos3theta - 3 sin3theta = Rcos(3theta + alpha)

    Expand the second equation, equate the co-efficients, cancel out the identical parts, then figure what R and alpha is from there.
    I know in this case you use Rcos, and what to do from there, but how did you know how to use Rcos in the first place instead of Rsin?
    Offline

    0
    ReputationRep:
    Because if you use the former equation you will not be able to cancel out the co-efficients as smoothly as you can with the latter equation?

    As far as i am aware, generally:

    if Asintheta + Bcostheta, then the equation Rsin (theta + alpha) must be applied.

    If Acostheta + Bsintheta, then the equation Rcos(theta + alpha) must be applied.

    as your equation starts with '2cos....' then the second rule is applied.

    hope that explains it
    • Thread Starter
    Offline

    0
    ReputationRep:
    but say you had the equation

    2sintheta + 3costheta = 5

    you could easily rearrange that to

    3costheta + 2sintheta = 5

    I dont understand why the order of them being different requires a completely different factor being taken out
    Offline

    0
    ReputationRep:
    (Original post by kej817)
    but say you had the equation

    2sintheta + 3costheta = 5

    you could easily rearrange that to

    3costheta + 2sintheta = 5

    I dont understand why the order of them being different requires a completely different factor being taken out
    Sorry i wrote the rules wrong...
    its actually (according to my textbook):


    Asintheta + Bcostheta, then the equation Rsin (theta + alpha) must be applied.

    If Acostheta + Bsintheta, then the equation Rcos(theta - alpha) must be applied.

    If Asintheta - Bcostheta, then the equation Rsin (theta - alpha) must be applied

    If Acostheta - Bsintheta, then the equation Rcos(theta + alpha) must be applied.

    Don't have time to explain it completely, but hopefully these rules can help you out!
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Student29)
    Sorry i wrote the rules wrong...
    its actually (according to my textbook):


    Asintheta + Bcostheta, then the equation Rsin (theta + alpha) must be applied.

    If Acostheta + Bsintheta, then the equation Rcos(theta - alpha) must be applied.

    If Asintheta - Bcostheta, then the equation Rsin (theta - alpha) must be applied

    If Acostheta - Bsintheta, then the equation Rcos(theta + alpha) must be applied.

    Don't have time to explain it completely, but hopefully these rules can help you out!
    Hey, thanks a lot, this will serve a lot of use for now anyway with my homework.
    I'll try and get my teacher to explain it to me why it works some time soon.
    Offline

    0
    ReputationRep:
    (Original post by Student29)
    Sorry i wrote the rules wrong...
    its actually (according to my textbook):


    Asintheta + Bcostheta, then the equation Rsin (theta + alpha) must be applied.

    If Acostheta + Bsintheta, then the equation Rcos(theta - alpha) must be applied.

    If Asintheta - Bcostheta, then the equation Rsin (theta - alpha) must be applied

    If Acostheta - Bsintheta, then the equation Rcos(theta + alpha) must be applied.

    Don't have time to explain it completely, but hopefully these rules can help you out!
    How is 'alpha' calculated for each of these equations?
    Offline

    0
    ReputationRep:
    Consider the addition formulae:

    \sin(\theta \pm \alpha) = \sin(\theta)\cos(\alpha) \pm \cos(\theta)\sin(\alpha)
    \cos(\theta \pm \alpha) = \cos(\theta)\cos(\alpha) \mp \sin(\theta)\sin(\alpha)

    What you're doing in these sorts of questions is just working backwards. :yep:

    Sorry, no time to elaborate at the moment; morning lecture.
    Offline

    2
    ReputationRep:
    (Original post by Student29)
    Sorry i wrote the rules wrong...
    its actually (according to my textbook):


    Asintheta + Bcostheta, then the equation Rsin (theta + alpha) must be applied.

    If Acostheta + Bsintheta, then the equation Rcos(theta - alpha) must be applied.

    If Asintheta - Bcostheta, then the equation Rsin (theta - alpha) must be applied

    If Acostheta - Bsintheta, then the equation Rcos(theta + alpha) must be applied.

    Don't have time to explain it completely, but hopefully these rules can help you out!
    Those rules seem pretty arbitrary, for example you don't need to write +alpha or -alpha. Just always write +alpha and sometimes you'll get a negative value.


    @OP Unless I'm missing something (if so, I only just got up so forgive me) but either one can work on any of the equations. After all every Cos graph is just a translated Sin graph etc.

    The exam itself should tell you which form to write it in surely?
    Offline

    1
    ReputationRep:
    by using tangent rule of opposite over adjacent and we know our opposite and adjacent as we found out what the coefficients of our equation were then do arctan of opposite over adjacent and you get alpha.
    Offline

    0
    ReputationRep:
    So it's arctan b/a in each case? Or a/b ? Can anyone be more specific?
    Offline

    1
    ReputationRep:
    (Original post by mathflair)
    So it's arctan b/a in each case? Or a/b ? Can anyone be more specific?
    arc tan 3/2
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.