# Answer to Question #4905 in Calculus for klinc

Question #4905

find the point on the curve y= cosx closest to the point (0,0)

Expert's answer

Distance formula:

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

Here, x1 = y1 = 0.

d = SQRT( x^2 + y^2 )

y = cos(x) - equation of our curve

d = SQRT( x^2 + (cos(x))^2 )

f(x) = x^2 + (cos(x))^2

f'(x) = 2x - 2cos(x)sin(x) = 2x - sin(2x)

Set f'(x) to zero:

2x - sin(2x) = 0

2x = sin(2x)

2x = 0, or x=0 - the unique solution that minimize the distance

cos(0) = 1

So, the point on the curve y=cosx closest to the point (0,0) is point (0,1).

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

Here, x1 = y1 = 0.

d = SQRT( x^2 + y^2 )

y = cos(x) - equation of our curve

d = SQRT( x^2 + (cos(x))^2 )

f(x) = x^2 + (cos(x))^2

f'(x) = 2x - 2cos(x)sin(x) = 2x - sin(2x)

Set f'(x) to zero:

2x - sin(2x) = 0

2x = sin(2x)

2x = 0, or x=0 - the unique solution that minimize the distance

cos(0) = 1

So, the point on the curve y=cosx closest to the point (0,0) is point (0,1).

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## Comments

Assignment Expert16.01.14, 17:22Take a look: the distance d = SQRT(x^2 + (cos(x))^2) and its square d^2=x^2 + (cos(x))^2 reaches the minimal value simultaneously. Thus one can find minimum of d^2 because it looks easier.

caroline werner15.01.14, 05:26how did you get rid of the SQRT after the 5th step? you can't just drop that trash....

Assignment Expert05.04.13, 14:242x = sin(2x)

Applying double angle formula we get

2x = 2sinx cosx

or

x=sin x cos x

This is a transcendental equation which can be solved graphically. This equation has unique solution x=0.

imran sabir04.04.13, 16:24how did you go from

2x = sin(2x)

to

2x = 0.

How six(2x) =0?

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