Results 1 to 7 of 7

Math Help - Precalculus Trig Problem

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    3

    Precalculus Trig Problem

    Hello,

    Here is a pre calculus problem I am having trouble with. The problem is from a review sheet for pre calculus, but I do not recall doing anything like this in my pre calculus class.

    (a) Express the function f(x) = - \sin 2x + \sqrt{3} \cos 2x in terms of sine only.

    (b) Find the amplitude A, period P, and phase shift B of the function, and graph one complete period.

    Any help is greatly appreciated.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jan 2009
    Posts
    91
    for part (a) use the identity  \cos 2x = 1 - 2 \sin^2 x
    and then you'll have it in terms of sin
    Last edited by Jhevon; January 29th 2009 at 06:31 PM. Reason: clarifying the identity
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by khodaja View Post
    Hello,

    Here is a pre calculus problem I am having trouble with. The problem is from a review sheet for pre calculus, but I do not recall doing anything like this in my pre calculus class.

    (a) Express the function f(x) = - \sin 2x + \sqrt{3} \cos 2x in terms of sine only.

    (b) Find the amplitude A, period P, and phase shift B of the function, and graph one complete period.

    Any help is greatly appreciated.

    Thanks
    What razorfever said for part (a) is fine, and you will get an answer in terms of sine that is correct, however, the form that you will get likely will not be one that helps you in part (b) a lot.

    alternatively, note that there is a formula that goes like this (i do not know what it's called, or if it even has a name), but it is based on the addition formula for sine.

    \boxed{A \sin x + B \cos x = C \sin (x + \theta)}

    here, A, ~B,~C, \text{ and } \theta are constants.

    basically, here (if we write 2 instead of x), A = -1, \text{ and } B = \sqrt{3} and we want to find C and \theta to have a nice form in sine that we can graph easily and find all the things we are required to find. first, we will modify the formula, to write 2x instead of just x.

    so here goes:

    C \sin (2x + \theta ) = C(\sin 2x \cos \theta + \sin \theta \cos 2x) .............by the addition formula for sine

    = \underbrace{C \cos \theta}_{\text{this is our }A} \sin 2x + \underbrace{C \sin \theta}_{\text{this is our }B} \cos 2x

    note that here we have

    C \cos \theta = -1 ...........(1)
    C \sin \theta = \sqrt{3} .........(2)

    thus we have two simultaneous equations, with two unknowns, we can solve for our unknowns. (try doing that yourself, it can come in handy to recall \sin^2 \theta + \cos^2 \theta = 1)

    anyway, there are infinitely many solutions here, any one will suffice. the nicest one i think is: C = 2, \theta = \frac {2 \pi}3

    thus, we have:

    f(x) = - \sin 2x + \sqrt{3} \cos 2x = 2 \sin \left( 2x + \frac {2 \pi}3\right)

    or in other words,

    f(x) = 2 \sin 2 \left( x + \frac {\pi}3 \right)

    can you take it from here?



    EDIT: in retrospect, it would perhaps have been easier to note the following:


    f(x) = - \sin 2x + \sqrt{3} \cos 2x

    = 2 \left( \underbrace{- \frac 12}_{\cos \frac {2 \pi}3} \sin 2x +  \underbrace{\frac {\sqrt{3}}2}_{\sin \frac {2 \pi}3} \cos 2x \right)

    = 2 \left( \sin 2x \cos \frac {2 \pi}3 + \sin \frac {2 \pi}3 \cos 2x \right)

    = 2 \sin \left( 2x + \frac {2 \pi }3\right)

    = 2 \sin 2 \left( x + \frac {\pi}3 \right)
    Last edited by Jhevon; January 29th 2009 at 09:06 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2009
    Posts
    3
    so, according to this solution, the amplitude is 2. right?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by khodaja View Post
    so, according to this solution, the amplitude is 2. right?
    yes

    what about the period and phase shift?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2009
    Posts
    3
    phase shift is : pi/3

    period is : pi
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by khodaja View Post
    phase shift is : pi/3
    not exactly.

    look back at the general form to see what the phase shift is

    period is : pi
    correct
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. precalculus problem
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 9th 2009, 01:31 AM
  2. Two Precalculus Trig Equations
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 29th 2009, 07:14 PM
  3. Precalculus Problem
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 2nd 2007, 04:10 PM
  4. Precalculus Problem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 13th 2007, 05:15 PM
  5. PRECALCULUS (Trig Functions) -SO LOST
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 30th 2006, 07:09 PM

Search Tags


/mathhelpforum @mathhelpforum