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Math Help - Trigonometric functions.

  1. #1
    Newbie Silvahere's Avatar
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    Trigonometric functions.

    Could someone clarify these questions step-by-step?
    Much appreciated! (sorry for my earlier thread!)

    Solve within [0, 2pi):

    cos2x + cosx = 0

    cotx - cosx = cotxcosx
    Last edited by Silvahere; November 27th 2009 at 01:02 PM.
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  2. #2
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    cotx - cosx = cotxcosx

    cos^2x/sin^2x - cos^2x=cos^2x/sin^2x cos^2x

    devide by cos^2x , => 1/sin^2x - 1 = cos^2x/sin^2x

    => 1-sin^2x = cos^2x

    = cos^2x+sin^2x = 1 => 1=1 ... i cant understand the point from this simplification !!! check again your question , the first one will be similar
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  3. #3
    Newbie Silvahere's Avatar
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    AHHHH!
    Wait wait wait, I'm so sorry!
    I posted incorrectly.
    It's supposed to be solved within [0, 2pi).
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  4. #4
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    Quote Originally Posted by Silvahere View Post
    Could someone clarify these questions step-by-step?
    Much appreciated! (sorry for my earlier thread!)

    Simplify the following:

    cos2x + cosx = 0

    cotx - cosx = cotxcosx
    you need to look at the directions for these two equations again.

    trig expressions are simplified ... trig equations are either identities to be verified or conditional equations to be solved.

    the first equation is a conditional equation.

    the second equation is an identity.
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  5. #5
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    Quote Originally Posted by Silvahere View Post
    Could someone clarify these questions step-by-step?
    Much appreciated! (sorry for my earlier thread!)

    Simplify the following:

    cos2x + cosx = 0

    cotx - cosx = cotxcosx
    cos(2x) = 2cos^2(x)-1

    2cos^2(x)-cos(x)-1=0. Solve using quadratic equation
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  6. #6
    Newbie Silvahere's Avatar
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    Ahhhhh! I'm sooo sorryyyy. x.x
    I posted my questions incorrectly, they're supposed to be solved within [0, 2pi). *Face palm*
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  7. #7
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    right the first one is easy to solve

    cos2x+cosx=0 => 2cos^2 (x) + cosx - 1 = 0

    => (cosx - 1 ) ( cosx +1/2 ) = 0

    => cosx=1 x=0 or cosx=-1/2 , x= pi - pi/3 , x= pi+pi/3 check the answers again
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  8. #8
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    the second identity is valid for any X ,
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  9. #9
    Newbie Silvahere's Avatar
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    I'm absolutely and completely klutzing out right now.
    Let me clarify again, because AGAIN, I made a mistake. >___<

    Okay.

    Solve the trigonometric functions within [0, 2pi).
    a. cos3x + cosx = 0

    b. cotx - cosx = cotxcosx
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  10. #10
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Silvahere View Post
    I'm absolutely and completely klutzing out right now.
    Let me clarify again, because AGAIN, I made a mistake. >___<

    Okay.

    Solve the trigonometric functions within [0, 2pi).
    a. cos3x + cosx = 0

    b. cotx - cosx = cotxcosx
    cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x)
    <br />
(2cos^2(x)-1)(cos(x))-2sin^2(x)cos(x) = 2cos^3(x)-cos(x)-2cos(x)+2cos^3(x))

    = 4cos^3(x)-3cos(x)


    \therefore \: \: 4cos^3(x)-2cos(x)= 2cos(x)(2cos^2(x)-1)=0
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  11. #11
    Newbie Silvahere's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x)
    <br />
(2cos^2(x)-1)(cos(x))-2sin^2(x)cos(x) = 2cos^3(x)-cos(x)-2cos(x)+2cos^3(x))

    = 4cos^3(x)-3cos(x)


    \therefore \: \: 4cos^3(x)-2cos(x)= 2cos(x)(2cos^2(x)-1)=0
    ... Uh... I understood up to: cos(3x) = cos(2x+x) = cos(2x)cos(x)... x_x

    Edit: Nevermind nevermind, I just got the edit. But where did the sines come from?
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  12. #12
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Silvahere View Post
    ... Uh... I understood up to: cos(3x) = cos(2x+x) = cos(2x)cos(x)... x_x
    That is the derivation. It is usually sufficient to learn that cos(3x)=4cos^3(x)-3cos(x)

    I just derived it because I don't know the triple angle formula
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