Question

Find an equation line of the tangent line to the given curve at the given point.

Answer 1

\[y=sinx \] at (pi/6 , 1)

Answer 2

y-1= cosx(x-(Pi/6))

Answer 3

find the slope of tangent first
y=sinx
diff it once
dy/dx =cosx
put x=pi/6
dy/dx = cos pi/6
=sqrt3 /2
tangent: y=sqrt3 /2 x +c
put x =pi/6 and y=1
1=(sqrt3 /2)(pi/6)+c
you can then find c
put the value you get into y=sqrt3 /2 x +c
We can then get the equation of tangent

Answer 4

Answer is : \[6\sqrt{3x} - 6y - \pi \sqrt{3} + 6 = 0\]