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

    2
    ReputationRep:
    Hi everyone,

    I have been asked to solve the equation in the attachment, between 0 and 2 pi. I have an answer of 5/12 pi (which is right), but I thought 19/12 pi would solve it, but it doesn't.

    Why is this - because the cos graph for radians says that it should?
    Attached Images
     
    Offline

    22
    ReputationRep:
    (Original post by Electrogeek)
    Hi everyone,

    I have been asked to solve the equation in the attachment, between 0 and 2 pi. I have an answer of 5/12 pi (which is right), but I thought 19/12 pi would solve it, but it doesn't.

    Why is this - because the cos graph for radians says that it should?
    We have \cos \left(x - \frac{\pi}{4}\right) = \sqrt{3}/2 so this means that we have since, sqrt(3)/2 is positive:

    \displaystyle x - \frac{\pi}{4} = \frac{\pi}{6} \, \text{or} \, 2\pi - \frac{\pi}{6}.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?
    Offline

    22
    ReputationRep:
    (Original post by Electrogeek)
    Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?
    Well... not quite, remember that if \pi/6 is a solution, then so is \pi/6 \pm 2\pi. This gives x outside 2pi and 0, so we ignore it.

    Buuut, x- \pi/4 = 2\pi - \pi/6 - 2\pi = -\pi/6 is also a solution...
    Offline

    22
    ReputationRep:
    (Original post by Electrogeek)
    Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?
    Whilst I've got a few minutes, I'll write up a thing.

    So you know that solutions to trig equations are given by using the CAST method, we take the arctrig function of the absolute value of the RHS, in this case: \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}.

    Now our solutions are given by the infinite list, we can split this into two infinite lists, one is:

    x - \frac{\pi}{4} = \frac{\pi}{6}, \frac{\pi}{6} + 2\pi, \frac{\pi}{6} - 2\pi, \frac{\pi}{6} + 4pi, \frac{\pi}{6} - 4pi, \frac{\pi}{6} \pm 2npi, \ldots

    And the other is:

    x - \frac{\pi}{4} = \underbrace{2\pi - \frac{\pi}{6}}_{11\pi/6}, \frac{11\pi}{6} \pm 2\pi, \frac{11\pi}{6} \pm 4\pi, \frac{11\pi}{6}\ pm 6pi, \ldots

    Side note: Getting the idea? The reason why this is so is because \cos \alpha is a 2\pi-periodic function so we have: \cos \alpha = \cos (\alpha \pm 2\pi) this is true for \sin \alpha as well - tangent is slightly different in the sense that it is a \pi-periodic function, so \tan \alpha = \tan (\alpha \pm \pi) and instead of adding \pm 2\pi to each solution, you'd add \pm \pi.

    Anyways, the first few solutions are x - \frac{\pi}{4} = -\frac{11\pi}{6}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6}, \frac{23\pi}{6}

    So adding \pi/4 to both sides and throwing anything that isn't in our range gives us: x = \frac{\pi}{12}, \frac{5\pi}{12}
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Well... not quite, remember that if \pi/6 is a solution, then so is \pi/6 \pm 2\pi. This gives x outside 2pi and 0, so we ignore it.

    Buuut, x- \pi/4 = 2\pi - \pi/6 - 2\pi = -\pi/6 is also a solution...
    Cool - I understand now. Thanks.
    Offline

    22
    ReputationRep:
    (Original post by Electrogeek)
    Cool - I understand now. Thanks.
    I typed up something above if you want to have a look.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    I typed up something above if you want to have a look.
    So... would these be right for these equations? (answers are in the attachment)
    Attached Images
     
    Offline

    22
    ReputationRep:
    (Original post by Electrogeek)
    So... would these be right for these equations? (answers are in the attachment)
    Assuming you mean between 0 and 2pi, I agree with the one for \tan but not for \sin you should have four solutions for the sin one.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Assuming you mean between 0 and 2pi, I agree with the one for \tan but not for \sin you should have four solutions for the sin one.
    Are the two given for sin right so far?

    Or is is -pi/12 (+/- pi), and 7/6 pi (+/- pi)?
    Offline

    22
    ReputationRep:
    (Original post by Electrogeek)
    Are the two given for sin right so far?

    Or is is -pi/12 (+/- pi), and 7/6 pi (+/- pi)?
    Should be 2x = 2\pi - \frac{\pi}{6}, 2\pi - \frac{\pi}{6} \pm 2\pi, \pi + \frac{\pi}{6}, \pi + \frac{\pi}{6} \pm 2\pi

    and then throw away everything you don't want.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Should be 2x = 2\pi - \frac{\pi}{6}, 2\pi - \frac{\pi}{6} \pm 2\pi, \pi + \frac{\pi}{6}, \pi + \frac{\pi}{6} \pm 2\pi

    and then throw away everything you don't want.
    Thank you! Makes a lot more sense now.
    Offline

    22
    ReputationRep:
    (Original post by Electrogeek)
    Thank you! Makes a lot more sense now.
    No problem.
 
 
 
  • 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
    Would you like to hibernate through the winter months?
    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.