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    Please Help with this.

    Solve: cosx = sinx + 0.1 using the substitution obtained from letting tan1/2x = t, 0<x<360.

    Thankyou for help. Will give rep.
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    (Original post by mather)
    Please Help with this.

    Solve: cosx = sinx + 0.1 using the substitution obtained from letting tan1/2x = t, 0<x<360.

    Thankyou for help. Will give rep.
    your rep wont count
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    If tan(x/2) = t, then you should be able to work out sin(x/2) and cos(x/2) in terms of t, and thus work out sin(x) and cos(x) in terms of t, then solve your equation for t and then get your x values.
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    (Original post by theone)
    If tan(x/2) = t, then you should be able to work out sin(x/2) and cos(x/2) in terms of t, and thus work out sin(x) and cos(x) in terms of t, then solve your equation for t and then get your x values.
    thanks, but i still don't quite understand. Wot's sinx/2 in terms of t? I think I can do it if I know.
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    tan(x/2)==t => sinx = 2t/(1+t); cosx = (1-t)/(1+t)

    => 1-t = 2t + 0.1(1+t)

    3.1 t = 0.9

    t = 0.29

    x/2 = arctan 0.29 = 16.17, 106.17

    x = 32.34, 212.34
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    (Original post by elpaw)
    tan(x/2)==t => sinx = 2t/(1+t); cosx = (1-t)/(1+t)

    => 1-t = 2t + 0.1(1+t)

    3.1 t = 0.9

    t = 0.29

    x/2 = arctan 0.29 = 16.17, 106.17

    x = 32.34, 212.34
    Thank you very very much.
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    (Original post by elpaw)
    tan(x/2)==t => sinx = 2t/(1+t); cosx = (1-t)/(1+t)
    Just to briefly expand this (and correct it i think):

    If t = tan(x/2) it follows that if we draw a right angled triangle with x/2 as one angle, and we call the opposite 't' and the adajcent '1' then the hypotenuse = root(1+t^2).

    So sin(x/2) = t/(root(1+t^2)) and cos(x/2) = 1/(root(1+t^2)) as such sin(x) = 2sin(x/2)cos(x/2) = 2t/(1+t^2) and cos(x) = cos^2(x/2) - sin^2(x/2) = (1-t^2)/(1+t^2).
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    (Original post by theone)
    Just to briefly expand this (and correct it i think):

    If t = tan(x/2) it follows that if we draw a right angled triangle with x/2 as one angle, and we call the opposite 't' and the adajcent '1' then the hypotenuse = root(1+t^2).

    So sin(x/2) = t/(root(1+t^2)) and cos(x/2) = 1/(root(1+t^2)) as such sin(x) = 2sin(x/2)cos(x/2) = 2t/(1+t^2) and cos(x) = cos^2(x/2) - sin^2(x/2) = (1-t)/(1+t^2).
    yeah, i thought there was a square there
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    Shame it doesn;t put up your tag lines.

    Have you read the Bible Code2 It too predicted the Twin Towers.
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    (Original post by Phil_C)
    Shame it doesn;t put up your tag lines.

    Have you read the Bible Code2 It too predicted the Twin Towers.
    yes i have read it it is a bit interesting. shame it onlhy predicts stuff after it has happened.
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    Please tell me this isnt edexcel P2 ... :s
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    (Original post by elpaw)
    yes i have read it it is a bit interesting. shame it onlhy predicts stuff after it has happened.
    i don't remeber saying that! i must have been well drunk last night.
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    It has predicted the some facts that are still to happen, Global war in 2006 centered on Israel and Palastinians. Both leaders are taking the book quite seriously. Rabin (sp?) assasination was also predicted and discussed with him indvance if it's occurance.

    Did you catch the Bible Code programme on TV last night?
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    (Original post by Phil_C)
    It has predicted the some facts that are still to happen, Global war in 2006 centered on Israel and Palastinians. Both leaders are taking the book quite seriously. Rabin (sp?) assasination was also predicted and discussed with him indvance if it's occurance.

    Did you catch the Bible Code programme on TV last night?
    Nah I missed it and I really wanted to see it. Tell me the gist of it. I caught the end of it when they said that the blokes original estimate of 1/62500 chance was actually 1/3 or sumthing like that...
 
 
 
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