What is the slope of the line normal to the tangent line of #f(x) = cscx-sin2x # at # x= (pi)/12 #?
1 Answer
After a long derivation, I got:
#f_N(pi/12) = [(2 - sqrt3)/(sqrt3(2 - sqrt3) + 2sqrt(2+sqrt3))]x + (12sqrt(2 - sqrt3) - 3(2 - sqrt3) + pisqrt(2+sqrt3))/(6(2 - sqrt3)) + (sqrt3pi)/12#
So the slope is
The tangent line shows as:
graph{(cscx - sin(2x) - y)(7.59218 - 16.1516x - y) = 0 [-6.2, 7.844, -2.215, 4.81]}
and you can verify that the slope is indeed the negative reciprocal of
From the following linearization formula:
#bb(f_T(a) = f(a) + f"'"(a)(x - a))#
we can get the equation of the tangent line. From that, the line normal to it is simply the one with the negative reciprocal slope, making it perpendicular, i.e. normal.
The derivative,
#f'(x) = -cscxcotx - 2cos2x#
So, at
#f'(pi/12) = -[csc(pi/12)cot(pi/12) + 2cos(pi/6)]#
and
#f(pi/12) = csc(pi/12) - sin(pi/6)#
so that
#f_T(pi/12) = f(pi/12) + f'(pi/12)(x - pi/12)#
#= csc(pi/12) - sin(pi/6) - [csc(pi/12)cot(pi/12) + 2cos(pi/6)](x - pi/12)#
Note that
#sin^2(x) = (1 - cos2x)/2 => sin^2(x/2) = (1 - cosx)/2#
#=> sin(x/2) = pmsqrt((1-cosx)/2)#
Similarly,
#=> cos(x/2) = pmsqrt((1+cosx)/2)# .
Since
This allows us to evaluate
#csc(pi/12) = 1/(sin(pi/12)) = 1/(sin(1/2*pi/6))#
#= 1/(sqrt((1 - cos(pi"/"6))/2)) = [...] = 2/sqrt(2 - sqrt3)#
Similarly,
#cot(pi/12) = cos(pi/12)/sin(pi/12) = [...] = sqrt(2+sqrt3)/sqrt(2 - sqrt3)#
So, we would then get the following tangent line:
#f_T(pi/12) = csc(pi/12) - sin(pi/6) - [csc(pi/12)cot(pi/12)(x - pi/12) + 2cos(pi/6)(x - pi/12)]#
Plug in our previous evaluations to get:
#= 2/sqrt(2 - sqrt3) - 1/2 - 2/sqrt(2 - sqrt3)sqrt(2+sqrt3)/sqrt(2 - sqrt3)(x - pi/12) - sqrt3(x - pi/12)#
#= (4 - sqrt(2 - sqrt3))/(2sqrt(2 - sqrt3)) - (2sqrt(2+sqrt3))/(2 - sqrt3)(x - pi/12) - sqrt3(x - pi/12)#
#= (4 - sqrt(2 - sqrt3))/(2sqrt(2 - sqrt3)) - (2sqrt(2+sqrt3))/(2 - sqrt3)x + (2sqrt(2+sqrt3))/(2 - sqrt3) pi/12 - sqrt3x + (sqrt3pi)/12#
#= (12 - 3sqrt(2 - sqrt3))/(6sqrt(2 - sqrt3)) + [-sqrt3 - (2sqrt(2+sqrt3))/(2 - sqrt3)]x + (pisqrt(2+sqrt3))/(6(2 - sqrt3)) + (sqrt3pi)/12#
#= [-sqrt3 - (2sqrt(2+sqrt3))/(2 - sqrt3)]x + (12sqrt(2 - sqrt3) - 3(2 - sqrt3) + pisqrt(2+sqrt3))/(6(2 - sqrt3)) + (sqrt3pi)/12#
#=> color(green)(f_T(pi/12) = stackrel("Slope")overbrace(-[(sqrt3(2 - sqrt3) + 2sqrt(2+sqrt3))/(2 - sqrt3)])x + stackrel("y-intercept")overbrace((12sqrt(2 - sqrt3) - 3(2 - sqrt3) + pisqrt(2+sqrt3))/(6(2 - sqrt3)) + (sqrt3pi)/12))#
Finally, when we take the negative reciprocal of the slope, we get the line normal to the tangent line:
#color(blue)(f_N(pi/12) = [(2 - sqrt3)/(sqrt3(2 - sqrt3) + 2sqrt(2+sqrt3))]x + (12sqrt(2 - sqrt3) - 3(2 - sqrt3) + pisqrt(2+sqrt3))/(6(2 - sqrt3)) + (sqrt3pi)/12)#
Or, the approximate decimal formula is:
#color(blue)(f_N(pi/12) ~~ 0.0619x + 7.59218)#