1. ## equation of tangent

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)

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

did you get this too?

2. 1. Are you sure about the function since $y\left(\frac{\pi}{2}\right) \neq 1$ ?

2. yes but no ! This is the equation of the tangent whereas you are asked the slope

3. I don't get the same answer for the first one.

$y=\sin(x)+\cos(2x)$

$y'=\cos(x)-2\sin(2x)$

$y-1=[\cos(\frac{\pi}{2})-2sin(\pi)](x-\frac{\pi}{2})$

There shouldn't be a slope. It is just a horizontal line. $y=1$. The point $(\frac{\pi}{2},1)$ is an extreme point.

I don't get the same answer for the first one.

$y=\sin(x)+\cos(2x)$

$y'=\cos(x)-2\sin(2x)$

$y-1=[\cos(\frac{\pi}{2})-2sin(\pi)](x-\frac{\pi}{2})$

There shouldn't be a slope. It is just a horizontal line. $y=1$. The point $(\frac{\pi}{2},1)$ is an extreme point.
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