Results 1 to 14 of 14

Math Help - Help solving trig equation 1

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    103

    Help solving trig equation 1

    my ugly attempt
    cos(2x-\pi/2) =\sqrt2/2
    cos2xcos\pi/2+sin2xsin\pi/2 = \sqrt2/2
    (2cos^2-1)(cos\pi/2)+(2sinxcosx)(sin\pi/2)=\sqrt2/2
    /edit got another step
    (2cos^2-1)(1)0+(2sinxcosx)(1)=\sqrt2/2
    (2sinxcosx)=\sqrt2/2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    recall that

    Quote Originally Posted by dorkymichelle View Post
    my ugly attempt
    cos(2x-\pi/2) =\sqrt2/2
    cos2xcos\pi/2+sin2xsin\pi/2 = \sqrt2/2
    (2cos^2-1)(cos\pi/2)+(2sinxcosx)(sin\pi/2)=\sqrt2/2
    /edit got another step
    (2cos^2-1)(1)0+(2sinxcosx)(1)=\sqrt2/2
    (2sinxcosx)=\sqrt2/2
    recall that cos\left(\frac{\pi}{2}\right) = 0
    and sin\left(\frac{\pi}{2}\right) = 1

    so

    cos2x(0)+sin2x(1) = \sqrt2/2

    can you finish from here
    Last edited by bigwave; December 8th 2009 at 08:47 PM. Reason: latex update
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    103
    I ended up with
    (2sinxcosx)=\sqrt2/2
    what is the next methodlical step?(i can't spell..)

    In the answers, it is 3\pi/8+n\pi,9\pi/8+n\pi (its a study guide for my next test)
    How can I solve an equation without a basic unit circle trig. value?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
    Quote Originally Posted by dorkymichelle View Post
    I ended up with
    (2sinxcosx)=\sqrt2/2
    what is the next methodlical step?(i can't spell..)

    In the answers, it is 3\pi/8+n\pi,9\pi/8+n\pi (its a study guide for my next test)
    How can I solve an equation without a basic unit circle trig. value?
    lets go back to sin\left(2x\right) = \frac{\sqrt{2}}{2}

    then

     <br />
sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}<br />

    therefore 2x = \frac{\pi}{4}

    so

    x = \frac{\pi}{8}

    if you plug this back in the original it will ck out

    if you add \pi to \frac{\pi}{8} you get \frac{9\pi}{8}
    n\pi where n is an interger
    this is because sin(x) has a period of \pi

    it is a unit circle value just not a commonly known one.
    Last edited by bigwave; December 8th 2009 at 11:39 PM. Reason: more info
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2009
    Posts
    103
    Thankyou!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Quote Originally Posted by bigwave View Post
    lets go back to sin\left(2x\right) = \frac{\sqrt{2}}{2}

    then

     <br />
sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}<br />

    therefore 2x = \frac{\pi}{4}

    so

    x = \frac{\pi}{8}

    if you plug this back in the original it will ck out

    if you add \pi to \frac{\pi}{8} you get \frac{9\pi}{8}
    n\pi where n is an interger
    this is because sin(x) has a period of \pi

    it is a unit circle value just not a commonly known one.
    Correct but you must also remember that sin is positive in the second quad at \frac{3\pi}{4}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2009
    Posts
    103
    hmm.. if I also use 3\pi/4 then x would = to 3\pi/8 +n\pi, that is on the answer, but \pi/8 +n\pi is not on the answer.
    And also, I don't understand how you get 9\pi/8?
    Did he just skip the 2st quadrant, and wrote the 2nd answer as going half a circle from the 2nd quadrant instead of using the answer from the 1st?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Quote Originally Posted by dorkymichelle View Post
    hmm.. if I also use 3\pi/4 then x would = to 3\pi/8 +n\pi, that is on the answer, but \pi/8 +n\pi is not on the answer.
    And also, I don't understand how you get 9\pi/8?
    Did he just skip the 2st quadrant, and wrote the 2nd answer as going half a circle from the 2nd quadrant instead of using the answer from the 1st?
    \frac{\pi}{8} +\pi n

    n just stands for an interger, n=1,2,3,4,...

    \frac{\pi}{8} +(\pi)(1)

    find common denominators

    \frac{\pi}{8} +\frac{8\pi}{8}= \frac{9\pi}{8}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Dec 2009
    Posts
    103
    So basicly, if I put \pi/8,3\pi/8 and 9\pi/8 they will all be correct answers?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Quote Originally Posted by dorkymichelle View Post
    So basicly, if I put \pi/8,3\pi/8 and 9\pi/8 they will all be correct answers?

    Yes since trig equations are periodic.

    The correct way to write your answer is

    \frac{\pi}{8} +\pi n, \frac{3\pi}{8} + \pi n

    You have to becareful some times because they may put a constraint on what your solutions may be.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Dec 2009
    Posts
    103
    The question told me to find all solutions
    The teacher's answer on the study guide is
    3\pi/8+n\pi and 9\pi/8 +n\pi
    so I guess both ways is correct, but my teacher is just picky about solutions?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Quote Originally Posted by dorkymichelle View Post
    The question told me to find all solutions
    The teacher's answer on the study guide is
    3\pi/8+n\pi and 9\pi/8 +n\pi
    so I guess both ways is correct, but my teacher is just picky about solutions?
    Correct they are both correct and will give you all the solutions
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Dec 2009
    Posts
    103
    Thankyou so much. i fainlly understand this. The differnt answers were confusing me day.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Quote Originally Posted by dorkymichelle View Post
    Thankyou so much. i fainlly understand this. The differnt answers were confusing me day.
    Glad I could help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: May 29th 2011, 01:37 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  3. Solving A Trig Equation
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: January 30th 2010, 12:02 PM
  4. Solving a trig equation
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 9th 2009, 07:08 PM
  5. Solving Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 21st 2007, 04:51 PM

Search Tags


/mathhelpforum @mathhelpforum