How do I find the closest point on a curve to (1,1)?

[Knight226]

Homework Statement

Find the point on the curve y = cos (x) closest to the point (1,1)

Homework Equations

Tangent line equation
y = f'(c)(x-c)+f(c)

Distance formula
d = SQRT( (x₁-x₂)^2 + (y₁-y₂)^2 )

The Attempt at a Solution

I have problem doing the f'(c) part. I get stuck when I have trigonometry in the equation where I have to set it to zero to find the critical number...

So far I have
d = SQRT( (x-1)^2 + (y-1)^2 )
y = cos (x)
d = SQRT( (x-1)^2 + (cos(x)-1)^2 )
*Here according to my textbook, what matter is what is inside radical sign*
f(x) = (x-1)^2 + (cos(x)-1)^2
f(x) = x^2-2x+2+cos^2x-2cosx

f'(x) = 2x - 2 - sin2x + 2sinx

Set f'(x) to zero
2x - 2 - sin2x + 2sinx = 0

Get stuck here :(
let say I do...
-sin2x + 2sinx = -2x + 2
then.. I don't know what to do next... I am horrible with trigonometry... Please help. Thank you very much in advance. Please let me know if I have made mistakes somewhere along the calculation.

 

[Matterwave]

The derivative of cos^2(x) is not -sin(2x), it's -2cos(x)sin(x) by the chain rule.

 

[Knight226]

Understood. Thank you! :D

 

[Mark44]

It's simpler if you work with the square of the distance, and minimize it. After all, if the square of the distance is minimized, then the distance will also be minimized.

D²(x) = (x - 1)² + (cos(x) - 1)²
d/dx[D²(x)] = 2(x - 1) + 2(cos(x) - 1)*(-sin(x))

Now set this derivative to 0 and find the critical points.

 

[Char. Limit]

Isn't 2cos(x)sin(x) equal to sin(2x) by some trig identity?

So I don't see what Matterwave finds false...

 

[Mark44]

Isn't 2cos(x)sin(x) equal to sin(2x) by some trig identity?

So I don't see what Matterwave finds false...

Nor do I. The identity you cited is the double angle formula for sin.

 

[Char. Limit]

Ok, I have a tendency to assume that if two people's facts differ, and I'm one of the people, the fault lies in my facts. Thanks for clearing that up.