Verified answer

f(x) = cos(x)

f'(x) = -sin(x)

Since we need to the tangent at π/4, plug in x = π/4 into f'(x) to get our tangent line's slope.

f'(x) = -sin(x)

==> f'(x) = -sin(π/4)

==> f'(x) = -√2/2

So we need a line that has a slope of -√2/2 and passes through (π/4,√2/2).

y = (-√2/2)x + b

==> √2/2 = (-√2/2)(π/4) + b

==> √2/2 = (-π√2)/8 + b

==> √2/2 + (π√2)/8 = b

==> b = (4√2)/8 + (π√2)/8

==> b = [ (4 + π)√2 ]/8

Therefore, the equation of the tangent line is:

y = (-√2/2)x + [ (4 + π)√2 ]/8

I have verified this on my TI-84 by graphing.

I hope that helps!