1. Are you sure about the function since ?
2. yes but no ! This is the equation of the tangent whereas you are asked the slope
Can someone please check my answers below for accurateness?
let me know if you get the same or different answer with me.
1. Find an equation of the tangent line to the curve y=sinx+cos2x at the point (pi/6, 1)
My final answer is:
y=[-(sqrt3)x]/2 + 1 + [(sqrt3) pi]/12
What did you get?
2. Find the slope of the tangent line to the curve y=cos(cosx) at x=pi/2
my final answer is y=1
did you get this too?
Thanks for your help!
and SO SORRY! typed question wrong in number 1, it is bolded now if you scroll up to check.. The point was actually (pi/6, 1).
so if this is the case, would my answer be correct now?
here is my work:
y' = cos (pi/6) - 2 sin (pi/3)
=-sqrt3/2
now y=mx+b I have to use this (says my teacher)
1=(-sqrt3/2) (pi/6) + b
isolate for b and I get
1+ (sqrt3/12)(pi) = b
therefore eqtn is
y=-sqrt3/2 x + 1+ (sqrt3/12) pi