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

1 Answer
Dec 20, 2016

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]}


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)#