# Thread: equation of the tangent line with trig

1. ## equation of the tangent line with trig

Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))

y'=cos[(sinx)(cosx)]
y'3π=cos[(sinx)(-1)]
cos [(0)(-1)]
cos(0)=1

y-0=1(x-3π)
y= x-3

2. Originally Posted by hazecraze
Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))

y'=cos[(sinx)(cosx)]
y'3π=cos[(sinx)(-1)]
cos [(0)(-1)]
cos(0)=1

y-0=1(x-3π)
y= x-3πWhat's wrong with my answer?

Your derivative's wrong: $(\sin({\sin x}))'=\cos({\sin x})\cos x$.

Tonio

3. Originally Posted by hazecraze
Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))

y'=cos[(sinx)(cosx)]

You wrote that wierd, so I'm not sure if you got it wrong, or just used bad notation.

y'3π=cos[(sinx)(-1)]
cos [(0)(-1)]
cos(0)=1

y-0=1(x-3π)
y= x-3πWhat's wrong with my answer?
The derivative is $cos(sin(x))cos(x)$.

$y'(3\pi)=cos(sin(3\pi))cos(3\pi)=cos(0)(-1)=-1$

4. Originally Posted by hazecraze
Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))

y'=cos[(sinx)(cosx)] <<<<<<<<< here. This line should read: y' = cos(sin(x)) * cos(x)
y'3π=cos[(sinx)(-1)]
cos [(0)(-1)]
cos(0)=1

y-0=1(x-3π)
y= x-3
The equation of the tangent line becomes now: $y = -x + 3\pi$