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    • Thread Starter

    Hiyas, I've done es question, but I'd be really gretful if some1 could check this to see if i've done it right or not. & if i've done it wrong, could i have some pointers plz? =) thanx for yout time!!

    Q) Find the equation of the tangent to the curve y=2sin(x - (pi/6) at the point where x= (pi/3)

    y=2sin(x - (pi/6)
    y= 2 * (sin (x - (pi/6))
    y = 2 (sinxcos(pi/6)) - 2(cosxsin(pi/6))
    y = 2sinx(root3/2) - 2cosx(1/2)
    y = (root3) sinx - cos x

    dy/dx = (root3) cos x + sin x

    At x=(pi/3), y= (root 3)sin(pi/3) - cos(pi/3)
    y= (root 3)(root3/2) - (1/2)
    y= (3/2) - (1/2)
    y= 1

    So point (pi/3, 1) lies on the tangent

    (& Gradient of tangent using derivative)

    dy/dx=(root3) cosx + sin x
    =(root3) cos(pi/3) + sin(pi/3)
    =(root3)(1/2) + (root3/2)
    =(root3/2) + (root3/2)

    staight line through (pi/3, 1) & M=(root 3)
    a b

    Eqn of tangent: y-b = m(x-a)
    y - 1 =(root3)(x - (pi/3))
    y - 1= (root3)x -((root3 pi)/3)
    (root3)x -((root3 pi)/3) -y + 1 =0

    Yeah thats what i got, but as I said I'm nae sure..nae any gd at these questions.

    Thanx 4 reading this!! xx

    That looks good, but one thing, you don't have to use trig identities to change the orginal function. You can differentiate it straight away, since

    d/dx[2sin(x-(pi/6))] = 2cos(x - (pi/6))

    And continue as normal. Should get the same answer with less work.
    • Thread Starter

    Thankku! lol, thought no1 would read through it & thanx 4 tha tip! =) xx
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