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

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.

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