Posted by **Bonaventutre** on Wednesday, March 27, 2013 at 1:19am.

Calculate the equation of the tangent and normal lines for the curve f(x) = tan 2x, at the point where the x-coordinate is: x = π/8.

- calculus -
**Reiny**, Wednesday, March 27, 2013 at 7:52am
dy/dx = 2sec^2 (2x)

when x = π/8

f(π/8) = tan π/4 = 1

so our point of contact is (π/8, 1)

slope of tangent = 2 sec^2 (π/4)

= 2(√2/1)^2 = 4

equation of tangent:

y - 1 = 4(x - π/8)

y - 1 = 4x - π/2

y = 4x + 1-π/2

normal has slope -1/4

y = (-1/4)x + b

but the point (π/8,1) lies on it, so

1 = (-1/4)(π/8) + b

b = 1 + π/32

y = (-1/4)x + 1+π/32

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