Results 1 to 4 of 4

Math Help - Solving and Proving Tri IDs

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    22

    Solving and Proving Tri IDs

    I'm given these two questions from a friend today, and i don't know how to solve them at all.

    4 cos4xsin4x+1=0
    where the range is -3pi ≤ x ≤ 2 pi


    All i know is that you're suppose to subsitute the 4x as x
    and thats basically it. any help?


    The second question i got was proofing Ids

    (Root(2)/2) ( cos pi/12 - sin pi/12) = 1/2


    How am i suppose to solve that?


    Thank you so much for helping me :]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by TeriyakiDonnQ View Post
    I'm given these two questions from a friend today, and i don't know how to solve them at all.

    4 cos4xsin4x+1=0
    where the range is -3pi ≤ x ≤ 2 pi


    All i know is that you're suppose to subsitute the 4x as x
    and thats basically it. any help?


    The second question i got was proofing Ids

    (Root(2)/2) ( cos pi/12 - sin pi/12) = 1/2


    How am i suppose to solve that?


    Thank you so much for helping me :]
    I'd use the double angle identity for sin

    \sin(2ax) = 2\sin(ax) \cos(ax)

    4\cos(4x) \sin(4x) +1 = 2\sin(8x) + 1 = 0

    \sin(8x) = -\frac{1}{2}

    8x = \arcsin \left(-\frac{1}{2}\right)

    x = \frac{1}{8} \arcsin \left(-\frac{1}{2}\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, TeriyakiDonnQ!

    The second question i got was proofing Ids

    . .  \frac{\sqrt{2}}{2}\left(\cos\frac{\pi}{12} - \sin\frac{\pi}{12}\right) \:=\: \frac{1}{2}

    How am i suppose to solve that?
    . .
    You don't solve it . . . you "proof" it.

    On the left, we have: . \frac{\sqrt{2}}{2}\cos\frac{\pi}{12} - \frac{\sqrt{2}}{2}\sin\frac{\pi}{12}

    . . Note that: . \sin\tfrac{\pi}{4} \:=\:\cos\tfrac{\pi}{4}\:=\:\tfrac{\sqrt{2}}{2}


    The expression becomes: . \sin\frac{\pi}{4}\cos\frac{\pi}{12} - \cos\frac{\pi}{4}\sin\frac{\pi}{12}


    And we have: . \sin\left(\frac{\pi}{4} - \frac{\pi}{12}\right) \;=\;\sin\frac{\pi}{6} \;=\;\frac{1}{2}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2010
    Posts
    22
    Thank youu e^(i*pi) and Soroban
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Steps for solving/proving trig identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 12th 2010, 05:53 PM
  2. Please help proving and solving
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 13th 2009, 01:55 PM
  3. Proving an identity that's proving to be complex
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 21st 2009, 02:30 PM
  4. Proving and then solving eqns
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: June 6th 2009, 06:08 AM
  5. Proving/Solving trig equations
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: April 19th 2009, 07:38 PM

Search Tags


/mathhelpforum @mathhelpforum