Find the equation of the tangent to the curve y = cos^2 x when x = pi/3 Your solution should use radian measure.

Question

i got the equation

y = 0.9996

is this correct? thanks

Answer 1

No, the slope is -0.866, your equation is also missing x and a constant :)

Answer 2

The slope of the tangent is the derivative of the curve at x = ^π/₃

You will need to use the chain rule to differentiate y = cos²x

Let u = cosx -> ^du/_dx = -sinx

Now y = u² -> ^dy/_du = 2u

^dy/_dx = ^dy/_du x ^du/_dx = 2u x (-sinx) = -2sinxcosx

Plug in x = ^π/₃ to get a slope of ^-√3/₂

Now calculate the y-coordinate of the point where x = ^π/₃ :

y = cos²(^π/3) = ¹/₄

Plug these values (x, y, and the slope) into the general equation for a straight line:

(y - ¹/₄) = ^-√3/₂ (x - ^π/₃)

y = ^-√3/₂ x + ^√3π/₆ + ¹/₄