Ok, I have this problem... Find an equation of the tangent line to the curve y= cos x where y= 0.50?

Question

the answer is suppose to look like this y = .# + .#
examples

y = -0.25 x + 1.03
y = -0.48 x + 1.12
y = -0.68 x + 1.24
y = -0.84 x + 1.38
y = -0.95 x + 1.50
or none of these

Answer 1

y = cos x

0.50 = cos x => x = pi/3

d/dx cos x = - sin x = d'(x)

d'(pi/3) = - sin (pi/3) = - 0.866

So your tangent line would have slope -0.866 so it looks like the answer is none of these.

Answer 2

First off, you need to find the slope of the tangent line at y = 0.5.

Keep in mind that when y = 0.5, y = 1/2, which means you have to solve for x when 1/2 = cos(x)

cos is equal to 1/2 at two points: pi/3 and 5pi/3

Therefore, we want to find the tangent lines at (pi/3,0.5) and (5pi/3, 0.5).

We calculate the derivative to find the slope of the tangent line.

y = cos(x)
y' = -sin(x)
Since we want the slope at x = pi/3 we have to plug pi/3 in for the derivative.

m1 = -sin(pi/3) = -sqrt(3)/2

Similarly, for x = 5pi/3

m2 = -sin(5pi/3) = -(-sqrt(3)/2) = sqrt(3)/2

Therefore,

(y - 0.5)/(x - pi/3) = sqrt(3)/2, meaning

y - 0.5 = sqrt(3)/2 x - (sqrt(3)/2) * (pi/3)
y = sqrt(3)/2 x - (sqrt(3)/2) * (pi/3) + 0.5

If we are dealing in exact values, it equals none of those.

Answer 3

Thats really simple...if the curve is cos x,then the gradient of the curve is -sinx as u differentiate it...frm the equation y=cosx,u find x..
cosx=0.50, x=60..
then,the equation of ur tangent is y=-sinx+c..u replac x by 60 n y by 0.50..
0.50=-sin60+c..
c=1.3660...n u have ur equation...