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# y=2sin(x - (pi/6) - tangents watch

1. 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)
=2(root3)/2
=(root3)

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.

2. 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.
3. Thankku! lol, thought no1 would read through it & thanx 4 tha tip! =) xx

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Updated: April 13, 2006
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