Question

consider the following.
Function
y = cos 3x
point (pi/4, -square root of 2/2)

(a) Find an equation of the tangent line to the graph of f at the given point. (Let x be the independent variable and y be the dependent variable.)

Answer 1

first, find the slope of the tangent line anywhere:

y' = -3sin(3x)
so, at x=π/4, y = -3sin(3π/4) = -3/√2

Now you have a point and a slope, so the line is

y + 1/√2 = -3/√2 (x - π/4)

Answer 2

y’= m (the slope) of the tangent line
y' = -3sin(3x)
so, at x=π/4, y = -3sin(3π/4) = -3(√2/2)

Then plug that into point slope formula: y-y1 = m (x-x1)

y-(-√2/2)= 3√2/2 (x- π/4)

I believe Steve’s work is right except that sin (3π/4)= √2/2, not 1/2