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    x=t+sint
    y=1-cost

    a) Show that √(x²+y²)=2cos½t

    I've gotten to
    RHS=2cost½t
    =2√(cos2t/2 + 1/2)

    LHS=√(t²+2tsint-2cost+1)

    But I'm not sure what to do after that
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    (Original post by bobbricks)
    x=t+sint
    y=1-cost

    a) Show that √(x²+y²)=2cos½t
    Perhaps I'm missing something but

    t=0 \Rightarrow x = 0 +\sin0 = 0, y = 1-\cos 0 = 1-1=0 \Rightarrow \sqrt{x^2+y^2} = 0

    and

    t=0 \Rightarrow 2\cos t/2 = 2 \cos 0/2 = 2

    so something doesn't look right here.
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    (Original post by atsruser)
    Perhaps I'm missing something but

    t=0 \Rightarrow x = 0 +\sin0 = 0, y = 1-\cos 0 = 1-1=0 \Rightarrow \sqrt{x^2+y^2} = 0

    and

    t=0 \Rightarrow 2\cos t/2 = 2 \cos 0/2 = 2

    so something doesn't look right here.
    I imagine its asking you to show it works for all values of t, and so you want to sub your x and y into the LHS and show it = RHS.
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    (Original post by SamKeene)
    I imagine its asking you to show it works for all values of t, and so you want to sub your x and y into the LHS and show it = RHS.
    How do you show LHS=RHS for all values of t (without using specific values of x and y)?
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    (Original post by bobbricks)
    x=t+sint
    y=1-cost

    a) Show that √(x²+y²)=2cos½t

    I've gotten to
    RHS=2cost½t
    =2√(cos2t/2 + 1/2)

    LHS=√(t²+2tsint-2cost+1)

    But I'm not sure what to do after that
    I got

    √(x²+y²)=t +sin½t

    i used sint = 2sin(t)/2cos(t/2)
    and cost = 1 - 2sin2(t/2) which leads to a perfect square

    is this any good?
 
 
 
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