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    sahill's Avatar
    sahill Posts: 1, Reputation: 1
    New Member
     
    #1

    Nov 11, 2004, 06:58 PM
    Trig Identities
    This may be a typo, but if not, is this solvable?
    1-cot^2theta=?
    CroCivic91's Avatar
    CroCivic91 Posts: 729, Reputation: 23
    Senior Member
     
    #2

    Dec 3, 2004, 03:01 AM
    1-cot^2theta = ?
    It is kind of important what you want this expression to be equal to...
    Lets assume you want it to be equal to 0...

    1-cot^2theta = 0 is equal to (1-cot(theta))*(1+cot(theta)) = 0
    Now, 2 numbers multiplied can produce 0 only if one (or both) of them are equal to 0.

    So, first lets assume that 1-cot(theta)=0
    That means that cot(theta) = 1 and that means that theta = pi/4 rad + k*pi, for any natural k

    Now, lets assume that 1+cot(theta)=0
    That means that cot(theta)=-1 and that means that theta = -pi/4 + k*pi, for any natural k

    Kresho
    Glipto's Avatar
    Glipto Posts: 7, Reputation: 1
    New Member
     
    #3

    Mar 9, 2005, 06:30 PM
    Hmmm... is not an ecuation.

    I think...

    Cos^2 theta + Sin^2 theta = 1 (Pitagoras)

    1 - Cos^2 theta = Sin^2 theta
    CroCivic91's Avatar
    CroCivic91 Posts: 729, Reputation: 23
    Senior Member
     
    #4

    Mar 10, 2005, 01:15 AM
    I would agree with you, but his original post said:

    1-cot^2theta=?

    cot != cos
    reinsuranc's Avatar
    reinsuranc Posts: 92, Reputation: 6
    Junior Member
     
    #5

    Mar 10, 2005, 11:10 AM
    Any chance that negative sign should be a positive sign?
    sin^2theta + cos^2theta = 1

    divide both sides by sin^2theta:

    (sin^2theta)/(sin^2theta) + (cos^2theta)/(sin^2theta) = 1/sin^2theta

    1 + cot^2theta = csc^2theta
    Glipto's Avatar
    Glipto Posts: 7, Reputation: 1
    New Member
     
    #6

    Mar 10, 2005, 09:56 PM
    OOPS!! I NEED GLASSES!!

    I agree with Reinsuranc, if the sign is positive the identity is beautiful.
    MathMaven53's Avatar
    MathMaven53 Posts: 20, Reputation: 2
    New Member
     
    #7

    Mar 23, 2005, 01:16 PM
    Trig Identity
    1- cot^2 t = 1 - cos^2 t/sin^2 t

    = (sin^2 t - cos^2 t)/sin^2 t

    = - cos 2t/sin^2 t


    Note that cos^2 t - sin^2 t = cos 2t

    2sin^2 t = 1 - cos 2t

    1- cot^2 t = - cos 2t/sin^2 t

    = (-2 cos 2t)/(2sin^2 t

    = (-2 cos 2t)/(1 - cos 2t)
    sammay_mon's Avatar
    sammay_mon Posts: 1, Reputation: 1
    New Member
     
    #8

    Jan 19, 2006, 01:57 PM
    Hay I'm new to this
    Do I ask a "?" and someone'll help me?

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