QUESTION Tangent - Trigonometry

[Brainiac]

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

 

[Ecom_Assistant]

To find the equation of the tangent line, we need to find the derivative of the function f(x) = ln(tan(2x)) and evaluate it at x = π/8.

First, let's find the derivative of f(x):
f'(x) = d/dx[ln(tan(2x))]
Using the chain rule, we have:
f'(x) = sec^2(2x) * 2
f'(x) = 2sec^2(2x)

Now, let's evaluate f'(x) at x = π/8:
f'(π/8) = 2sec^2(2(π/8))
f'(π/8) = 2sec^2(π/4)
f'(π/8) = 2(1)
f'(π/8) = 2

So, the slope of the tangent line at x = π/8 is 2.

To find the equation of the tangent line, we use the point-slope form:
y - y1 = m(x - x1)

Using the point (π/8, ln(tan(2(π/8)))), we have:
y - ln(tan(2(π/8))) = 2(x - π/8)

Simplifying, we get:
y - ln(tan(π/4)) = 2x - π/4

The equation of the tangent line is:
y = 2x - π/4 + ln(tan(π/4))

To find the equation of the normal line, we use the fact that the slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is -1/2.

Using the point-slope form again, we have:
y - ln(tan(2(π/8))) = -1/2(x - π/8)

Simplifying, we get:
y - ln(tan(π/4)) = -1/2x + π/16

The equation of the normal line is:
y = -1/2x + π/16 + ln(tan(π/4))

I hope this helps! Let me know if you have any further questions.

 

[Mr.Admin]

Woa! very details