Question

What is the slope of the line normal to the tangent line of `f(x) = cscx-secx` at `x= (17pi)/12`?

Answer 1

First off, you may want to refer to this answer as additional practice.

Furthermore, we will be using a similar method to the one in the answer linked above, and we'll also reference information that we've already figured out in that answer.

I got the line normal to the tangent line as:

`f_N((17pi)/12) = -[1/(2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2")]x + 1/6 [12sqrt(2 + sqrt3) - 12sqrt(2 - sqrt3) - 17(2 - sqrt3)^"3/2"pi - 17(2 + sqrt3)^"3/2"pi]`

The slope of this normal line is therefore `-[1/(2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2")]`.

Or the approximate decimal normal line is:

`f_N((17pi)/12) ~~ -0.0680x - 62.5816`

You can see the tangent line itself here:

graph{(cscx - secx - y)(14.6969x - 62.5816 - y) = 0 [2.392, 6.833, -0.24, 5.306]}

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Linearization Equation

`bb(f_T(a) = f(a) + f"'"(a)(x - a))`

where `a = (17pi)/12`.

Since this isn't half of a convenient angle, we'll end up using the addition formulas later for `sin` and `cos`, in addition to the half-angle relations we already figured out in a previous answer.

For now:

`f'(x) = -(cscxcotx + secxtanx)`
`f(x) = cscx - secx`

so that

`f'((17pi)/12) = -[csc((17pi)/12)cot((17pi)/12) + sec((17pi)/12)tan((17pi)/12)]`

`f((17pi)/12) = csc((17pi)/12) - sec((17pi)/12)`

Since `cscx = 1/sinx` and `secx = 1/cosx`, we can use the addition formulas to evaluate these trig functions directly. Notice how `(17pi)/12` radians is also `(3pi)/2 - pi/12`.

Now, recall from here that:

`sin(pi/12) = sqrt((1 - cos(pi"/"6))/2) = [...] = sqrt(2 - sqrt3)/2`
`cos(pi/12) = sqrt((1 + cos(pi"/"6))/2) = [...] = sqrt(2 + sqrt3)/2`

Also, recall the addition formulas:

`sin(u pm v) = sinucosv pm cosusinv`
`cos(u pm v) = cosucosv ∓ sinusinv`

Thus:

`color(green)(sin((17pi)/12)) = sin((3pi)/2 - pi/12)`

`= sin((3pi)/2)cos(pi/12) - cancel(cos((3pi)/2)sin(pi/12))^(0)`

`= color(green)(-sqrt(2 + sqrt3)/2)`

`color(green)(cos((17pi)/12)) = cos((3pi)/2 - pi/12)`

`= cancel(cos((3pi)/2)cos(pi/12))^(0) + sin((3pi)/2)sin(pi/12)`

`= color(green)(-sqrt(2 - sqrt3)/2)`

This allows us to evaluate the following functions:

• `csc((17pi)/12) = 1/sin((17pi)/12) = -2/sqrt(2 + sqrt3)`
• `cot((17pi)/12) = cos((17pi)/12)/sin((17pi)/12) = sqrt(2 - sqrt3)/sqrt(2 + sqrt3)`
• `sec((17pi)/12) = 1/cos((17pi)/12) = -2/sqrt(2 - sqrt3)`
• `tan((17pi)/12) = 1/cot((17pi)/12) = sqrt(2 + sqrt3)/sqrt(2 - sqrt3)`

Plugging all of these in to `f'(a)` and `f(a)`, we get:

`color(green)(f'((17pi)/12)) = -[-2/sqrt(2 + sqrt3)sqrt(2 - sqrt3)/sqrt(2 + sqrt3) - 2/sqrt(2 - sqrt3)sqrt(2 + sqrt3)/sqrt(2 - sqrt3)]`

`= -[-(2sqrt(2 - sqrt3))/(2 + sqrt3) - (2sqrt(2 + sqrt3))/(2 - sqrt3)]`

`= (2(2 - sqrt3)^"3/2")/cancel((2 + sqrt3)(2 - sqrt3))^(1) + (2(2 + sqrt3)^"3/2")/cancel((2 - sqrt3)(2 + sqrt3))^(1)`

`= color(green)(2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2")`

`color(green)(f((17pi)/12)) = -2/sqrt(2 + sqrt3) + 2/sqrt(2 - sqrt3)`

`= -(2sqrt(2 - sqrt3))/cancel(sqrt(2 + sqrt3)sqrt(2 - sqrt3))^(1) + (2sqrt(2 + sqrt3))/cancel(sqrt(2 - sqrt3)sqrt(2 + sqrt3))^(1)`

`= color(green)(2sqrt(2 + sqrt3) - 2sqrt(2 - sqrt3))`

Plugging these results into the linearization equation, we can obtain the tangent line equation to simplify.

`f_T((17pi)/12) = stackrel(f(a))overbrace(2sqrt(2 + sqrt3) - 2sqrt(2 - sqrt3)) + stackrel(f'(a))overbrace([2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2"])(x - (17pi)/12)`

Multiply through:

`= 2sqrt(2 + sqrt3) - 2sqrt(2 - sqrt3) + 2(2 - sqrt3)^"3/2"x + 2(2 + sqrt3)^"3/2"x - 2(2 - sqrt3)^"3/2"(17pi)/12 - 2(2 + sqrt3)^"3/2"(17pi)/12`

Get common denominators going on:

`= [2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2"]x + (12sqrt(2 + sqrt3))/6 - (12sqrt(2 - sqrt3))/6 - (17pi)/6 (2 - sqrt3)^"3/2" - (17pi)/6 (2 + sqrt3)^"3/2"`

Combine fractions with common denominators:

`= [2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2"]x + (12sqrt(2 + sqrt3) - 12sqrt(2 - sqrt3))/6 - [(17pi)/6 (2 - sqrt3)^"3/2" + (17pi)/6 (2 + sqrt3)^"3/2"]`

`=> color(green)(f_T((17pi)/12) = [2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2"]x + 1/6 [12sqrt(2 + sqrt3) - 12sqrt(2 - sqrt3) - 17(2 - sqrt3)^"3/2"pi - 17(2 + sqrt3)^"3/2"pi])`

So, the line normal to the tangent line will have the negative reciprocal slope:

`color(blue)(f_N((17pi)/12) = -[1/(2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2")]x + 1/6 [12sqrt(2 + sqrt3) - 12sqrt(2 - sqrt3) - 17(2 - sqrt3)^"3/2"pi - 17(2 + sqrt3)^"3/2"pi])`

Or the approximate decimal answer is:

`color(blue)(f_N((17pi)/12) ~~ -0.0680x - 62.5816)`