Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Trigonometry help.

  1. Feb 7, 2010 #1
    1.
    Hi
    Its the bit (c) where i am stuck at although it doesn't look much complicated, for all i know is that max value for cos =1 , so cos inverse becomes 0. The answer on the mark scheme is theta = 326 which i cant figure out.Thanks
    2. Relevant equations

    10i5wcw.png


    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 7, 2010 #2

    I'm just wondering, have you tried it? I would start by observing that

    [tex]cos(\theta + \alpha) = cos(\theta) cos(\alpha) - sin(\theta) sin(\alpha)[/tex]

    After solving for [tex]\alpha[/tex], you then can find a maximum [tex]cos(\theta+\alpha)[/tex] which when multiplied by R will give the answer to part (c).

    If you need further help, post your work so far.
     
  4. Feb 7, 2010 #3
    Ah how weird i just did another similar question of which i got the answers hmm anyway, here is what i did:
    i got the R value square root. 13 (by equating co-efficients of sin and cos),
    theta 33.7.
    P.S: Sorry i have yet to become latex friendly , if only someone could post me a tutorial on how to use it lol .
     
  5. Feb 7, 2010 #4
    Note that [tex]cos(\theta+\alpha)=1[/tex] means that [tex]\theta=-\alpha[/tex]. However, the pattern repeats every [tex]360^o[/tex].
     
  6. Feb 7, 2010 #5
    Oh! would i let R[tex]cos(\theta+\alpha)[/tex]= 1?
     
  7. Feb 7, 2010 #6
    No problem. A good start with LaTex is http://frodo.elon.edu/tutorial/tutorial/". Others can be found by Googling "latex tutorial". At any rate, take care, 73s and clear skies.
     
    Last edited by a moderator: Apr 24, 2017
  8. Feb 7, 2010 #7
    Thanks alot!:) for the link .
     
    Last edited by a moderator: Apr 24, 2017
  9. Feb 7, 2010 #8
    Not quite. Let [tex]cos(\theta+\alpha)=1[/tex]. That will give the maximum for [tex]3 cos(\theta) - 2 sin(\theta)[/tex] after it is multiplied by R. Since R is effectively a constant any maximum of [tex]cos(\theta+\alpha)[/tex] will be proportionate to [tex]R cos(\theta+\alpha)=3 cos(\theta) - 2 sin(\theta)[/tex].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook